Open tubes

In Chapter 31, “The Compact-Open Topology”, it is asserted that “clearly” the third topology on \Mor(X, Y) (with X and Y the real line) is finer than the second. The issue is more fully discussed here.

In this post I discuss the concept of open tubes, particular subsets of the Cartesian product X \times Y with X a set and Y a metric space.

Definition

Let X be a set and Y a metric space, with distance d.

For any r > 0 and any a \in Y, we note \dot B(a, r) the open ball of Y of radius r and centred on a, that is set of all x \in Y such that d(a, x) < r. Open balls are notoriously open subsets.

Let \phi be any mapping X \to Y, and r a real > 0.

Then the open tube \dot T(\phi, r) of radius r and centred on \phi is the following subset of X \times Y:

\dot T(\phi, r) = \{\ (x, y) \in X \times Y\ |\ d(\phi(x), y) < r\ \}

Equivalently:

\dot T(\phi, r) = \bigcup_{x \in X} \{x\} \times \dot B(\phi(x), r)

The “open” in “open tube” refers to its being based on open balls (second expression above), but does not imply that it is itself necessarily open; indeed, we have not given any topology to X, so we don’t have any particular one on X \times Y and cannot speak of \dot T(\phi, r) being open or not.

However, if X is a topological space and \phi is continuous X \to Y, then \dot T(\phi, r) is open, as I will show.

An open tube centred on a continuous mapping is open

If X is a topological space and \phi is a continuous mapping X \to Y, then any open tube centred on \phi is an open subset of X \times Y.

Proof:

I will show that in the above circumstances \dot T(\phi, r) is a neighbourhood of each of its points.

Let (x_0, y_0) be an eventual point of \dot T(\phi, r).

Then d(\phi(x_0), y_0) < r.

Lettuce define s = \frac 1 2 (r - d(\phi(x_0), y_0)). We have s > 0.

Let D = \phi^\leftarrow(\dot B(\phi(x_0), s)). Since \phi is continuous and \dot B(\phi(x_0), s) is an open subset of Y, D is an open subset of X. Since \phi(x_0) \in \dot B(\phi(x_0), s), x_0 \in D.

Let O = D \times \dot B(y_0, s), open subset of X \times Y. Clearly, (x_0, y_0) \in O, so O is a neighbourhood of (x_0, y_0).

Let (x, y) be an eventual point of O.

x \in D, hence \phi(x) \in \dot B(\phi(x_0), s), that is d(\phi(x), \phi(x_0)) < s.

y \in \dot B(y_0, s), hence d(y_0, y) < s.

d(\phi(x), y) \le d(\phi(x), \phi(x_0)) + d(\phi(x_0), y_0) + d(y_0, y)

Since d(\phi(x), \phi(x_0)) < s and d(y_0, y) < s, this leads to:

d(\phi(x), y) < s + d(\phi(x_0), y_0) + s

With:

s + d(\phi(x_0), y_0) + s = 2 s + d(\phi(x_0), y_0)
\ = 2 (\frac 1 2 (r - d(\phi(x_0), y_0))) + d(\phi(x_0), y_0)
\ = r

Thus we have: d(\phi(x), y) < r.

Which implies: (x, y) \in \dot T(\phi, r).

Hence O \subseteq \dot T(\phi, r).

Thus we have found a neighbourhood O of (x_0, y_0) that is a subset of \dot T(\phi, r), which implies that \dot T(\phi, r) itself is a neighbourhood of (x_0, y_0).

Thus \dot T(\phi, r) is an open subset of X \times Y. \blacksquare

Uniform convergence topology and the open set topology

Chapter 31, “The Compact-Open Topology”, describes the compact-open topology, and goes on to compare it with two other topologies on \Mor(X, Y) in the case where X = Y = \R. It asserts that “clearly” the second of these three is finer than the first, and the third finer than the second.

This may well be “clear” intuitively, but is not so easy to prove. Here I will prove that the third (that the author does not name, but that I call the open set topology) is finer than the second, which is the uniform convergence topology.

I will only need to specify that X is a topological space and Y a metric space with distance d.

I define E = \Mor(X, Y) the set of continuous mappings from X to Y.

I note \dot B(y, r) the open ball in Y centred on y \in Y and with radius r > 0. It is the set of all y' \in Y such that d(y, y') < r.

Furthermore, for any H subset of X \times Y, I note \kappa(H) the set of all continuous mappings \phi: X \to Y the graph of which is a subset of H; in other words, \kappa(H) = \{\ \phi \in E\ |\ \forall x \in X, (x, \phi(x)) \in H\ \}.

The uniform convergence topology on E is such that a subset A of E is open if and only if for any \phi \in A, there exists r > 0 such that for any \psi \in E, if \forall x \in X, \psi(x) \in \dot B(\phi(x), r), then \psi \in A. We can express this condition in terms of subsets of X \times Y that I have named open tubes: for any mapping (not necessarily continuous) \phi: X \to Y and for any r > 0, the open tube \dot T(\phi, r) is \{\ (x, y) \in X \times Y\ |\ d(\phi(x), y) < r\ \}. Then A \subseteq E is open for the uniform convergence topology if and only if for any \phi \in A, there exists r > 0 such that \kappa(\dot T(\phi, r)) \subseteq A.

The third topology ‑ the “open set” topology ‑ is the topology on E generated by all the \kappa(O) for all O open subsets of X \times Y. (The book says the topology is this collection itself, but I have not managed to prove that it directly forms a topology ‑ as is more explicitly requested in exercise 215.)

To show that the open set topology is finer than the uniform convergence one, we consider an open set for the uniform convergence topology and endeavour to show that it is open in the open set one.

Let A be an open set in the uniform convergence topology.

Let \phi be an eventual member of A.

There exists an r > 0 such that \kappa(\dot T(\phi, r)) \subseteq A.

I prove here that, \phi being continuous, \dot T(\phi, r) is an open subset of X \times Y. It includes the graph of \phi. Hence \kappa(\dot T(\phi, r)) contains \phi, and is by definition an open set for the open set topology on E. Since it is a subset of A, A is a neighbourhood of \phi for the open set topology.

Hence for any \phi \in A, A is a neighbourhood of \phi for the open set topology. This implies that A is itself an open set in the open set topology. \blacksquare

Connected subsets one of which meets the closure of the other

Exercise 224 (Chapter 32, “Connectedness”):

Let A and B be connected subsets of topological space X such that A intersects \Cl(B). Prove that A \cup B is connected. Find an example to show that it is not enough to assume that \Cl(A) intersects \Cl(B).

Let us assume that V and W are open subsets of X such that V \cap W \cap (A \cup B) is empty and that A \cup B \subseteq V \cup W. To show that A \cup B is connected, we need to show that either V or W does not intersect A \cup B.

Since A is connected, and V \cap W \cap A is empty and A \subseteq V \cup W, necessarily A is entirely in V or in W. Similarly, B is entirely in V or entirely in W.

Let us suppose that they are not both in V or both in W; for instance, that A \subseteq V and B \subseteq W.

If a \in A, then a \in V with V open in X. Since V does not intersect B, a cannot be in \Cl(B). Hence A cannot intersect \Cl(B). This contradicts our assumption.

Hence it is impossible that A \subseteq V and B \subseteq W. Similarly, it is impossible that A \subseteq W and B \subseteq V. Thus, A and B are either both in V or both in W, which implies that either V or W does not intersect A \cup B. \blacksquare

If we only assume that \Cl(A) and Cl(B) intersect, it does not always follow that A \cup B is connected. For instance, let X be the topological plane and A and B two open disks that “almost touch”, that is such that the corresponding closed disks ‑ which are their closures ‑ intersect in one point. The union of A and B is then not connected.

Union of transitively overlapping connected spaces

Exercise 225 (Chapter 32, “Connectedness”):

Prove the following generalization of theorem 40. Let A_\lambda (\lambda in \Lambda \neq \emptyset) be a collection of connected subsets of topological space X, and let the equivalence relation on this collection of sets generated by A_\lambda \approx A_{\lambda'} if A_\lambda \cap A_{\lambda'} \neq \emptyset have just one equivalence class. Then \bigcup_\lambda A_\lambda is connected.

For clarity, rather than a family (A_\lambda)_{\lambda \in \Lambda} I will consider a set \mathcal A of connected subsets of X. The condition is that the equivalence relation in \mathcal A generated by the above relation \approx has only one class.

Let K = \bigcup_{A \in \mathcal A} A.

Let V and W be eventual open subsets of X such that K \subseteq V \cup W and V \cap W \cap K = \emptyset. To show that K is connected, we must show that either V \cap K or W \cap K is empty.

First, let A be an eventual element of \mathcal A. Since A \subseteq K, the two subsets of A, A \cap V and A \cap W are disjoint, and their union is A. Furthermore, they are open in A. Since A is connected, this implies that one or the other is empty. If A \cap V is empty, then A \subseteq W; if instead A \cap W is empty, then A \subseteq V. Hence in all cases an element of \mathcal A is entirely in V or entirely in W.

Let us consider the relation \mathcal R on \mathcal A defined by A_1 \mathcal R A_2 \iff (A_1 \subseteq V \wedge A_2 \subseteq V) \vee (A_1 \subseteq W \wedge A_2 \subseteq W); in other words, A_1 \mathcal R A_2 if and only if both are subsets of V, or both are subsets of W. Clearly, this is an equivalence relation.

Furthermore, if A_1 \approx A_2, that is, A_1 \cap A_2 \neq \emptyset, we cannot have A_1 \subseteq V and A_2 \subseteq W, for V and W are disjoint. Hence A_1 and A_2 are either both in V or both in W; that is, A_1 \mathcal R A_2. Hence \mathcal R is an equivalence relation greater than the equivalence relation generated by \approx.

If \mathcal R had two or more classes, there would be some element of \mathcal A in one class and another in another; these two elements would not be equivalent following \mathcal R. Since \mathcal R is greater than the equivalence relation generated by \approx, they would not be equivalent either following this latter relation; which contradicts the fact that this has only one class. Hence \mathcal R too can have only one class, which means that for all A_1 and A_2 in \mathcal A, either both are in V or both are in W. But this implies that all are in V or all are in W, which in turn implies that their union K is a subset of either V or W, hence that eitherV \cap K or W \cap K is empty. \blacksquare

Direct products/sums: when the associated morphisms are epi/mono

Exercise 7 (Chapter 2, “Categories”):

In the category of sets, the two morphisms (\alpha and \beta) in a direct product are monomorphisms and the two morphisms in a direct sum are epimorphisms. Is this true in every category?

There are two mistakes in this wording.

  • The author has exchanged “monomorphisms” and “epimorphisms”. In a direct product of sets, the morphisms are generally not monomorphisms (injective), and in a direct sum of sets, they are generally not epimorphisms (surjective).
  • In a direct product of sets, the morphisms are almost always epimorphisms. If one of the sets is empty, the product is empty, and the morphism associated with the other set (an empty mapping) is not an epimorphism; unless this other set is empty too.

The general theorem I will prove is this:

Let A and B be objects of a category \mathfrak C, and (P, \alpha, \beta) a \mathfrak C-direct product of A and B. If \Mor(A, B) is not empty then \alpha is an epimorphism.

We consider the above conditions satisfied. There exists a morphism \gamma: A \to B.

Let \alpha' be the identity morphism A \to A and \beta' the morphism \gamma: A \to B. Applying the universal property of the direct product (P, \alpha, \beta) to the triplet (A, \alpha', \beta'), we obtain that there exists a unique \epsilon: A \to P such that both \alpha' = \alpha \circ \epsilon and \beta' = \beta \circ \epsilon.

The first of these two relations says that \alpha \circ \epsilon is the identity on A. If we have an object X and two morphisms \phi_1 and \phi_2 from A to X such that \phi_1 \circ \alpha = \phi_2 \circ \alpha, then, composing this relation on the right with \epsilon and eliminating the resulting identity morphisms, we get \phi_1 = \phi_2. Hence \alpha is an epimorphism. \blacksquare

We obtain, of course, a similar result exchanging the roles of A and B.

Reversing the arrows, we get the corresponding result for direct sums:

Let A and B be objects of a category \mathfrak C, and (S, \alpha, \beta) a \mathfrak C-direct sum of A and B. If \Mor(B, A) is not empty then \alpha is a monomorphism.

Plus, of course, the similar result obtained by exchanging A and B.

It so happens that in the category of sets, in the case of a direct product, both \alpha and \beta are always monomorphisms, even if \Mor(B, A), respectively \Mor(A, B), are empty.

Theorems about the direct product of two topological spaces

A lot of results concerning the direct product (Z, \alpha, \beta) of two topological spaces X and Y are trivial if you consider the usual particular construction of these entities. I find it more satisfying to derive them directly from the universal definitions.

See also: the equivalent page on direct sums.

In the following, we will always have X and Y topological spaces and ((Z, \tau), \alpha, \beta) a direct product thereof.

No duplicates

In the Z = X \times Y construction, this is just the trivial fact that if you know (a, b), you know both a and b.

Theorem:

For every c_1, c_2 \in G, if \alpha(c_1) = \alpha(c_2) and \beta(c_1) = \beta(c_2), then c_1 = c_2.

Proof:

Let us suppose that we have c_1, c_2 \in Z such that \alpha(c_1) = \alpha(c_2) and \beta(c_1) = \beta(c_2).

Let us take Z' a singleton topological space, that is that Z' is a singleton (singletons exist!) and we equip it with the only possible topology, which is both the indiscrete and the discrete one.

We define \alpha' and \beta' as the constant mappings, Z' \to X and Z' \to Y respectively, that map the only element of Z' to \alpha(c_1), which is also \alpha(c_2), and to \beta(c_1), which is also \beta(c_2), respectively.

We define \gamma_ and \gamma_2 as the constant mappings, both Z' \to G, that map the only element of Z' to c_1 and to c_2 respectively.

Since these four mappings are constant, or, alternatively, since the topology on Z' is discrete, they are all continuous.

Furthermore, for both \gamma = \gamma_1 and \gamma = \gamma_2 we have:

    \[\alpha' = \alpha \circ \gamma\]

    \[\beta' = \alpha \circ \gamma\]

Since (Z, \alpha, \beta) is a direct product of X and Y, there can be only one morphism \gamma: Z' \to G satisfying these two relations. Hence necessarily \gamma_1 = \gamma_2, which implies c_1 = c_2. \blacksquare

Existence of an antecedent

If we have a \in X and b \in Y, does there exist a c \in Z such that \alpha(c) = a and \beta(c) = b? Again, this is trivial in the Z = X \times Y representation: c = (a, b) is the answer. Sticking to the universal definition of the direct product, we have:

For any a \in X and b \in Y, there exists c \in Z such that \alpha(c) = a and \beta(c) = b.

Proof:

We take Z' as a singleton topological space as above, and \alpha' and \beta' as the mappings Z' \to X and Z' \to Y respectively that map the unique element of Z to a and to b respectively. These two mappings again are continuous, since Z' has the discrete topology.

The universal definition of (Z, \alpha, \beta) as a direct product implies that there exists \gamma: Z' \to Z such that \alpha' = \alpha \circ \gamma and \beta' = \alpha \circ \gamma. Taking c as the image by \gamma of the unique element of Z, we thus have a = \alpha(c) and b = \beta(c). \blacksquare.

This result and the preceding one come together as:

For any a \in X and b \in Y, there exists a unique c \in Z such that \alpha(c) = a and \beta(c) = b.

The associated morphisms are surjective

At least, usually.

If X = \emptyset or Y \neq \emptyset, then \alpha is surjective. If Y = \emptyset or X \neq \emptyset, then \beta is surjective.

This is just a consequence of the general result proved in exercise 7 (corrected): in any category, if (P, \alpha, \beta) is a direct product of A and B, and if \Mor(A, B) \neq \emptyset, then \alpha is an epimorphism (and similarly, exchanging the roles of A and \alpha, and B and \beta, respectively); plus the fact that in the category of topological spaces, epimorphisms are surjective mappings.

Characterisation of the topology

That the topology on the the direct product space Z is the one generated by the collection of all \alpha^\leftarrow(U) and \beta^\leftarrow(V) for U open in X and V open in Y is a stipulation of the explicit construction of Z as X \times Y. It can however be proven without reference to this construction.

The topology of Z is the topology generated in Z by the collection of all \alpha^\leftarrow(U) and \beta^\leftarrow(V) for U open in X and V open in Y. A subset of Z is open if and only if it is the union of some collection of subsets of the form \alpha^\leftarrow(U) \cap \beta^\leftarrow(V) with U open in X and V open in Y.

Proof:

Let \tau be the direct product topology on Z and \tau_0 the topology on Z generated in Z by the collection of all \alpha^\leftarrow(U) and \beta^\leftarrow(V) for U open in X and V open in Y.

Since \alpha and \beta are defined as continuous, necessarily \tau contains all the \alpha^\leftarrow(U) and \beta^\leftarrow(V) for U open in X and V open in Y. Hence \tau is finer than \tau_0. We wish to show that \tau_0 is also finer than \tau.

The mapping \alpha, considered from (Z, \tau_0) to X, remains continuous, since all the \alpha^\leftarrow(U) for U open in X are in \tau_0. Similarly, \beta is continuous (Z, \tau_0) \to Y. Hence we have a topological space (Z, \tau_0) and two continuous mappings \alpha and \beta, from (Z, \tau_0) to X and Y respectively. Since ((Z, \tau), \alpha, \beta) is a direct product of X and Y, there exists a unique continuous mapping \gamma: (Z, \tau_0) \to (Z, \tau) such that \alpha = \alpha \circ \gamma and \beta = \beta \circ \gamma.

This implies that any element c \in Z has the same image as \gamma(c) by both \alpha and \beta. The first theorem above implies that \gamma(c) = c. Thus \gamma = \Id_Z.

Hence \Id_Z is continous (Z, \tau_0) \to (Z, \tau); which means that \tau_0 is finer than \tau.

Hence \tau = \tau_0, that is the topology \tau of the direct product is the topology of Z generated by the collection of all \alpha^\leftarrow(U) and \beta^\leftarrow(V) for U open in X and V open in Y.

The open sets of the topology generated by a collection of sets \mathcal B are the unions of any number of finite intersections of elements of \mathcal B.

In this case, \mathcal B = \{ \alpha^\leftarrow(U) \}_{ U\ open\ in\ X } \cup \{ \beta^\leftarrow(V) \}_{ V\ open\ in\ Y }. A finite intersection of elements of \mathcal B will be of the form \bigcap_{i=1}^p \alpha^\leftarrow(U_i) \cap \bigcap_{j=1}^q \beta^\leftarrow(V_j) with the U_i‘s open subsets of X and the V_j‘s open subsets of Y. Since intersections are well-behaved respective to inverse images, and since finite intersections of open subsets are open subsets, this can be written \alpha^\leftarrow(\bigcap_{i=1}^p U_i) \cap \beta^\leftarrow(\bigcap_{j=1}^q V_j) = \alpha^\leftarrow(U) \cap \beta^\leftarrow(V) with U open subset of X and V open subset of Y. Hence an open subset of Z is a union of subsets of Z of the form \alpha^\leftarrow(U) \cap \beta^\leftarrow(V) with U open subset of X and V open subset of Y\blacksquare

Continuity of a mapping to a product space

If E is a topological space and \phi a mapping E \to Z, then \phi is continuous if and only if both \alpha \circ \phi and \beta \circ \phi are continuous.

If \phi: E \to Z is continuous, since both \alpha: Z \to X and \beta: Z \to Y are continuous, by composition of continuous mappings we conclude that \alpha \circ \phi and \beta \circ \phi are continuous.

Conversely, suppose that we know that \alpha \circ \phi and \beta \circ \phi are continuous. Let W be an eventual open subset of Z. We wish to show that \phi^\leftarrow(W) is an open subset of E.

Let a be an eventual point of \phi^\leftarrow(W). Then \phi(a) \in W. Since W is open in Z and (see above) “a subset of Z is openif and only if it is the union of some collection of subsets of the form \alpha^\leftarrow(U) \cap \beta^\leftarrow(V) with U open in X and V open in Y”, there exist U open in X and V open in Y such that \phi(a) \in \alpha^\leftarrow(U) \cap \beta^\leftarrow(V) \subseteq W. Taking \phi^\leftarrow of this expression, we obtain a \in \phi^\leftarrow(\alpha^\leftarrow(U) \cap \beta^\leftarrow(V)) \subseteq \phi^\leftarrow(W). But \phi^\leftarrow(\alpha^\leftarrow(U) \cap \beta^\leftarrow(V)) = \phi^\leftarrow(\alpha^\leftarrow(U)) \cap \phi^\leftarrow(\beta^\leftarrow(V)) = (\alpha \circ \phi)^\leftarrow(U) \cap (\beta \circ \phi)^\leftarrow(V). Since we know that both \alpha \circ \phi and \beta \circ \phi are continuous, this last expression is an open subset of E, that contains a and is included in \phi^\leftarrow(W). Thus \phi^\leftarrow(W) is a neighbourhood of a. This being the case for any a \in \phi^\leftarrow(W), \phi^\leftarrow(W) is an open subset of E. \blacksquare

Direct sum of Hausdorff topological spaces

Exercise 185 (Chapter 28, “The category of topological spaces”):

Prove that both the direct product and the direct sum of two Hausdorff topological spaces is Hausdorff.

We will consider Hausdorff topological spaces X and Y and their direct sum (in the first part) or product (in the second part) (Z, \alpha, \beta).

We will proceed without reference to the explicit constructions of direct topological sums and products, using instead the results in “Theorems about the direct product and sum of two topological spaces”.

Direct sum

Let c_1 and c_2 be two eventual points of Z, with c_1 \neq c_2. The first result about direct sums in the above page implies that \alpha^\rightarrow(X) \cap \beta^\rightarrow(Y) = \emptyset. The second result implies that \alpha^\rightarrow(X) \cup \beta^\rightarrow(Y) = Z. Two cases are possible:

  • One of c_1 and c_2 is in \alpha^\rightarrow(X) and the other in \beta^\rightarrow(Y) .
  • Both c_1 and c_2 are in \alpha^\rightarrow(X), or both are in \beta^\rightarrow(Y).

We will consider successively the case where c_1 \in \alpha^\rightarrow(X) and c_2 \in\beta^\rightarrow(Y), and the case where both are in \alpha^\rightarrow(X). The other possible cases are similar and will lead to similar results.

First case (c_1 \in \alpha^\rightarrow(X) and c_2 \in\beta^\rightarrow(Y)):

We have c_1 = \alpha(a) and c_2 = \beta(b), with a \in X and b \in Y.

X itself is an open neighbourhood of a, and hence (last result about direct sums on the above page) \alpha^\rightarrow(X) is an open neighbourhood of c_1 = \alpha(a) in Z. Similarly, \beta^\rightarrow(Y) is an open neighbourhood of c_2 in Z. Since \alpha^\rightarrow(X) and \beta^\rightarrow(Y) are disjoint, we have found disjoint neighbourhoods respectively of c_1 and c_2.

Second case (c_1, c_2 \in \alpha^\rightarrow(X)):

We have c_1 = \alpha(a_1) and c_2 = \alpha(a_2), with a_1, a_2 \in \alpha^\rightarrow(X). Since c_1 \neq c_2, we have a_1 \neq a_2. Since X is Hausdorff, there exist U_1 and U_2 disjoint open neighbourhoods in X respectively of a_1 and a_2.

//TODO (we need \alpha injective)

Direct product

Let c_1 and c_2 be two eventual points of Z, with c_1 \neq c_2.

The first result about direct products in the above page implies that  \alpha(c_1) \neq \alpha(c_2) or \beta(c_1) \neq \beta(c_2).

Let us suppose that \alpha(c_1) \neq \alpha(c_2).

Since X is Hausdorff and \alpha(c_1) and \alpha(c_2) are distinct points of X, there exist open sets U_1 and U_2 of X such that \alpha(c_1) \in U_1, \alpha(c_2) \in U_2 and U_1 \cap U_2 = \emptyset.

Then \alpha^\leftarrow(U_1) and \alpha^\leftarrow(U_2) are disjoint (the image by \alpha of a common element would be both in U_1 and U_2, which are disjoint). They are open (since U_1 and U_2 are open and \alpha is continuous). Since \alpha(c_1) \in U_1, we have c_1 \in \alpha^\leftarrow(U_1); similarly, c_2 \in \alpha^\leftarrow(U_2). Thus there exist two disjoint open neighbourhoods respectively of c_1 and c_2 in Z.

The same would result if we had \beta(c_1) \neq \beta(c_2).

This being the case for all c_1, c_2 \in Z with c_1 \neq c_2, the topological space Z is Hausdorff.

Theorems about the direct sum of two topological spaces

A lot of results concerning the direct sum (Z, \alpha, \beta) of two topological spaces X and Y are trivial if you consider the usual particular construction of these entities. I find it more satisfying to derive them directly from the universal definitions.

See also: the equivalent page on direct products.

In the following, we will always have X and Y topological spaces and ((Z, \tau), \alpha, \beta) a direct sum thereof.

Non-overlapping images

\alpha^\rightarrow(X) and \beta^\rightarrow(Y) are disjoint.

Let a and b be eventual elements respectively of X and Y.

Let k and l be any two distinct objects (such exist), and Z' = \{ k, l \}, equipped with any topology (the discrete one will do). Then the constant mappings \alpha': X \to Z', x \mapsto k and \beta': Y \to Z', y \mapsto l are continuous.

Hence there exists a continuous mapping \gamma: Z \to Z' such that \alpha' = \gamma \circ \alpha and \beta' = \gamma \circ \beta.

This implies in particular that \alpha'(a) = \gamma(\alpha(a)) and \beta'(b) = \gamma(\beta(b)), that is, \gamma(\alpha(a)) = k and \gamma(\beta(b)) = l. Since k \neq l, it follows that \alpha(a) \neq \beta(b).

Thus \alpha^\rightarrow(X) and \beta^\rightarrow(Y) cannot have a common point, which would both be the image by \alpha of a point a of X and by \beta of a point b of Y.

The images cover the direct sum

\alpha^\rightarrow(X) \cup \beta^\rightarrow(Y) = Z

Let Z' = \alpha^\rightarrow(X) \cup \beta^\rightarrow(Y) with the topology induced on Z' \subseteq Z by that of Z.

Let \alpha' be the mapping X \to Z', x \mapsto \alpha(x) and \beta' the mapping Y \to Z', y \mapsto \beta(y).

Since an open set U' on Z' is the intersection of Z' with an open set U on Z and since Z' \backslash Z certainly contains no points of \alpha^\rightarrow(X), its preimage by \alpha' is equal to the preimage of U by \alpha and hence is open in X. Thus \alpha' is continuous. Similarly, \beta' is continuous.

We can thus apply the universal property of the direct sum (Z, \alpha, \beta), which implies that there exists a continuous mapping \gamma': Z \to Z' such that \alpha' = \gamma' \circ \alpha and \beta' = \gamma' \circ \beta.

Let \gamma be the mapping Z \to Z, z \mapsto \gamma'(z). The preimage by \gamma of an open set of Z is again the same as its preimage by \gamma', hence is an open set of Z. Hence \gamma is continuous.

For any x \in X, \alpha(x) = \alpha'(x) = (\gamma' \circ \alpha)(x) = \gamma'(\alpha(x)) = \gamma(\alpha(x)); hence \alpha = \gamma \circ \alpha. Similarly, \beta = \gamma \circ \beta.

But we also have \alpha = \Id_Z \circ \alpha and \beta = \Id_Z \circ \beta, with \Id_Z continuous Z \to Z. The uniqueness requirement of the universal property of the direct sum (Z, \alpha, \beta) thus implies that \gamma = \Id_Z.

But the image of \gamma is that of \gamma', and is included in Z'. Thus the image of \Id_Z, which is Z, is included in Z'. It follows that Z = Z', that is, that \alpha^\rightarrow(X) \cup \beta^\rightarrow(Y) = Z.

Topology on the direct sum

The topology on the direct sum Z is the topology generated by the collection of the \alpha^\rightarrow(U) and \beta^\rightarrow(V) for all U open in X and all V open in Y.

//TODO

Universal definition of the indiscrete topology

Exercise 182 (Chapter 29, “The category of topological spaces”):

Give a universal definition which leads to the introduction of the indiscrete topology on a set.

On a set X, the indiscrete topology is the unique topology such that for any set Y, any mapping \phi: Y \to X is continuous.

This isn’t really a universal definition. We have seen that the discrete topology \tau_D can be defined as the unique topology \tau that makes ((X, \tau), \Id_X) a free topological space on the set X. We will give the indiscrete topology, which is at the other end of the spectrum of topologies, a similar definition.

For any category \mathfrak C, we can define a “reversed arrows” category \mathfrak C^R, which has the same objects as \mathfrak C and the same morphisms, except that the sources and targets of the morphisms are reversed. Composition too is, obviously, reversed; the identities are the same. It is easily checked that this does lead to a proper category, and also that the arrows-reversed category of an arrows-reversed category leads back to the original category.

If we have categories \mathfrak C and \mathfrak D and a covariant functor \mathcal F from \mathfrak D to \mathfrak C, then we can consider the three modified functors \mathcal F^{DR} from \mathfrak D to \mathfrak C^R, \mathcal F^{RD} from \mathfrak D^R to \mathfrak C and\mathcal F^{RR} from \mathfrak D^R to \mathfrak C^R. These functors will map both objects and morphisms exactly as the original \mathcal F does. Those with only one reversal are contravariant; the one with a reversal in both categories is covariant, and is the one that will interest us here.

We consider the classical category of sets \catSet; in its arrows-reversed version, \catSet^R, the morphisms from set A to set B are the mappings from B to A. Similarly, we have the category of topological spaces \catTop and its arrows-reversed version \catTop^R in which morphisms from topological space (X, \tau) to topological space (Y, \sigma) are the continuous mappings from(Y, \sigma) to(X, \tau).

For a given set A, object of \catSet^R, we can ask if we can find a free object (X, \tau) in category \catTop^R following the forgetful functor \mathcal F^{RR} from \catTop^R to \catSet^R. This mens finding a topological space (X, \tau) and a \catSet^R-morphism \alpha: A \to \mathcal F^{RR}((X, \tau)) = X such that for any topological space (Y, \sigma) and any \catSet^R-morphism \alpha': A \to \mathcal F^{RR}((Y, \sigma)) = Y, there exists a unique \catTop^R-morphism \gamma: (X, \tau) \to (Y, \sigma) such that (as \catSet^R-morphisms) \alpha' = \gamma \circ \alpha.

Factoring in the arrows-reversals, this means that we wish to find a topological space (X, \tau) and a mapping \alpha: X \to A such that for any topological space (Y, \sigma) and any mapping \alpha': Y \to A, there exists a unique continuous mapping \gamma: (Y, \sigma) \to (X, \tau) such that (as mappings) \alpha' = \alpha \circ \gamma.

We propose that the topological space (A, \tau_I) with \tau_I the indiscrete topology on A, together with the mapping \alpha = \Id_A, is a free object on A following the functor \mathcal F^{RR}.

Proof:

Let (Y, \sigma) be a topological space and \alpha' a mapping Y \to A. If \gamma is to be a continuous mapping (Y, \sigma) \to (A, \tau_I) such that \alpha' = \alpha \circ \gamma, it must be a mapping Y \to A such that \alpha' = \Id_A \circ \gamma = \gamma; that is, we must have \gamma = \alpha'. This \gamma is indeed continuous, since \tau_I is the indiscrete topology. Hence it is the unique continuous mapping (Y, \sigma) \to (A, \tau_I) such that \alpha' = \alpha \circ \gamma.

Thus we have shown that A with the indiscrete topology, together with the identity mapping on A, is a free topological space following the arrows-reverse categorical definition of free objects.

Free objects are unique up to an isomorphism. However, we wish to show more specifically that given the set A, the indiscrete topology \tau_I on A is the unique topology \tau such that ((A, \tau), \Id_A) be a free object on A following the above definition. We could use the “unique up to an isomorphism” card, but more simply: If ((A, \tau), \Id_A) is to be a free object on A following the above definition, in particular, taking (Y, \sigma) = (A, \tau_I) and \alpha' = \Id_A, we need there to exist a continuous \gamma: (Y, \sigma) \to (A, \tau), that is, (A, \tau_I) \to (A, \tau) such that \alpha' = \Id_A \circ \gamma, that is such that \gamma = \Id_A. In other words, we need \Id_A to be continuous from (A, \tau_I) to (A, \tau), which implies \tau \subseteq \tau_I, that is, that \tau must be coarser than \tau_I; which is possible only if \tau = \tau_I.

Hence the indiscrete topology on A is the unique topology \tau such that ((A, \tau), \Id_A) be a free object in category \catTop^R following the functor \mathcal F^{RR} from category \catTop^R to category \catSet^R.

Representation frameworks and representations

Chapter 22 on representations is, I feel, the worst written I have come upon up to now. I haven’t yet come to wrap my head around it.

I certainly appreciate the desire to give a general definition of representations in category theory terms. This is how the author puts it:

Let \mathfrak C and \mathfrak C' be two categories. We suppose that we are given the following two things: i) a forgetful functor \mathcal F from the category \mathfrak C to the category of sets, and ii) a rule which assigns, to each object P' in category \mathfrak C', an object Z in category \mathfrak C and an isomorphism (in the category of sets) from set \mathcal F(Z) to the set \Mor(P', P').

I see two issues about this setting. I will “correct” these issues in this post, giving what I hope is a better definition, and come to the issue of direct products and sums of representations and of subrepresentations in another post.

About the functor to the category of sets

About the above complex statement, the author reassures us:

There is an obvious forgetful functor from every category we shall consider to the category of sets, and we shall always use this one for item i). The purpose of i) is to allow us to speak of “elements” of objects in category \mathfrak C.

The existence of this “obvious functor” in the categories we consider is very familiar; indeed, groups, vector spaces and so on are built upon sets. We are clearly used to speaking of an element of a group, though in fact we should speak of an element of the underlying set.

Postulating the existence of this forgetful functor \mathcal F from \mathfrak C to \catSet is, however, not enough to ensure that we are in the “familiar situation”. There is another element of this familiar situation that is not captured by the mere existence of this functor. Namely: in the familiar situation, the morphisms between objects of \mathfrak C are no more and no less than morphisms – i.e., mappings – between the corresponding sets. If O_1 and O_2 are objects of \mathfrak C, then \Mor(O_1, O_2) is literally a subset of \Mor(\mathcal F(O_1), \mathcal F(O_2)). This implies in particular that \mathcal F is “one to one” in its action on morphisms; there cannot be two different morphisms \phi, \psi: O_1 \to O_2 such that \mathcal F(\phi) = \mathcal F(\psi).

That this is not necessarily implied by the mere existence of a functor \mathcal F from \mathfrak C to \catSet can be made clear by considering this trivial fact: for any category \mathfrak C, there is a functor \mathcal F_0 from \mathfrak C to \catSet that takes any object O of \mathfrak C to the set \{O\} and any morphism in \mathfrak C between objects O_1 and O_2 to the one and only mapping that exists from set \{O_1\} to set \{O_2\}. One can check that this is indeed a functor; but clearly, in general it will take many different morphisms in \mathfrak C to just one morphism in \catSet.

To be closer to the familiar situation, we thus need (at least) the following additional condition: that the functor \mathcal F from \mathfrak C to \catSet be “morphism-injective”, that is, such that, for any given objects O_1 and O_2 of \mathfrak C and any \phi, \psi \in \Mor(O_1, O_2), we have \mathcal F(\phi) = \mathcal F(\psi) \implies \phi = \psi.

We will see why this condition is necessary particularly in the case of the direct products and sums of representations. In the book, it is hidden behind the notations of the author, that tend to muddle the distinction between morphisms in \mathfrak C and the corresponding morphisms in \catSet.

Must it really be an isomorphism?

The author postulates that there must be “ii) a rule which assigns, to each object P' in category \mathfrak C', an object Z in category \mathfrak C and an isomorphism (in the category of sets) from set \mathcal F(Z) to the set \Mor(P', P')“.

It makes for a simpler formulation if instead of speaking of an isomorphism, we say that this rule \mathcal R must assign to the object P' of \mathfrak C' an object Z of \mathfrak C such that \mathcal F(Z) = \Mor(P', P').

For instance: if \mathfrak C' is the category of real vector spaces, and \mathfrak C that of real associative algebras, we want our rule \mathcal R to “make the set of endomorphisms of a given real vector space P' into an associative algebra”, that is, \mathcal R(P') = Z is an associative algebra the underlying set of which is \Mor(P', P').

But this doesn’t work well if, instead, we want \mathfrak C to be the category of groups. The author calls this a “special case of the definition of a representaton”, but doesn’t clearly note that in this case, we much change his definition, replacing “isomorphism” by “monomorphism”. In the case of groups, it is not the whole of \Mor(P', P') that is “made into a group”, but only the set of bijective linear mappings from P' to itself.

So we will adopt the modified specification of ii): We need a rule \mathcal R which assigns, to each object P' in category \mathfrak C', an object \mathcal R(P') = Z such that \mathcal F(Z) \subseteq \Mor(P', P).

My definition of a representation framework

A representation framework is made of:

  • Two categories \mathfrak C and \mathfrak C'.
  • A morphism-injective functor from \mathcal C to \catSet.
  • A rule \mathcal R that assigns to each object P' in category \mathfrak C', an object \mathcal R(P') = Z such that \mathcal F(Z) \subseteq \Mor(P', P).

A representation is…

Given such a representation framework, a representation of an object A of \mathfrak C is an ordered pair (P', \psi) where P' is an object of \mathfrak C' and \psi a \mathcal C-morphism A \to \mathcal R(P').

This is very close to the definition given by the author. He then goes on to state:

(…) for each element a of object A, we must have a certain morphism, which we write \psi_a, from object P' to itself.

 

This “\psi_a” is misleading, in that it gives the impression that \psi_a is the image of a by \psi. But \psi is a morphism between A and Z = \mathcal R(P'). It is not (in general) a mapping at all. It is not \psi, but \mathcal F(\psi) that is a mapping and can map a to this \psi_a, element of \mathcal F(Z) \subseteq \Mor(P', P').

Despite the awkwardness, I will often prefer the notation “\mathcal F(\psi)(a)” to the shorter “\psi_a“, because the latter can give rise to the notion that if we happen to have, for each a, a certain \mathfrak C'-morphism P' \to P' that we call \gamma_a, for instance, then necessarily this mapping forms a \mathfrak C-morphism \gamma: A \to Z; while it only means that we have a mapping of sets \mathcal F(A) \to \mathcal F(Z) ‑ if indeed the \gamma_a‘s are all in \mathcal F(Z), which may be only a subset of \Mor(P', P') ‑ and this mapping may or may not be some \mathcal F(\gamma) with \gamma a \mathfrak C-morphism from A to Z. This false reasoning arises precisely in the presentations by the author of direct products of representations, direct sums and subrepresentations. That will be for another post.