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.

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.

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}.


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.

Identity functors, isofunctors and equivalent categories

Exercise 105 (Chapter 17, “Functors”):

Define the identity functor from a category to that same category. What do you suppose is meant by equivalent categories?

Identity functors

We have encountered two flavors of identity definitions. Identity mappings on sets are defined by the way they act on set elements, namely that they don’t change them. Identities in abstract categories, on the other hand, are defined by the way they behave respective to the composition of morphisms.

Categories are a bit like sets, in that they have “elements”, or rather two kinds of such: objects and morphisms. Inspired by this, we can define identity functors in a category \mathcal C as the functor \mathcal I from itself to itself that maps each object to itself and each morphism to itself. Such a functor always exists, and, since its effect is completely specified, it is unique.

But since we also have composition of functors, we can also try to define the identity functor on category \mathcal C as a functor, if it exists, \mathcal I_\circ from \mathcal C to \mathcal C such that for any category \mathcal C', for any functor \mathcal F from \mathcal C to \mathcal C', \mathcal F \circ \mathcal I_\circ = \mathcal F and for any functor \mathcal F' from \mathcal C' to \mathcal C, \mathcal I_\circ \circ \mathcal F' = \mathcal F.

It is immediate that the functor \mathcal I from the first definition satisfies the second. Since the former always exists, so does the latter. Furthermore, if there were two identity functors following the second definition, \mathcal I_\circ and \mathcal I'_\circ, we would have both \mathcal I_\circ \circ \mathcal I'_\circ = \mathcal I_\circ and \mathcal I_\circ \circ \mathcal I'_\circ = \mathcal I'_\circ, hence \mathcal I'_\circ = \mathcal I_\circ.

Thus the two definitions are equivalent; one and only one identity functor exists for each category.

Isofunctors and equivalent categories

The same exercise asks us to define the notion of equivalent categories.

We may first want to define an isofunctor between two categories:

A functor \mathcal F from category \mathcal C to category \mathcal C' is an isofunctor if and only if there exists a functor \mathcal F' from \mathcal C' to \mathcal C such that \mathcal F' \circ \mathcal F is the identity functor on \mathcal C and \mathcal F \circ \mathcal F' is the identity functor on \mathcal C'.

We can then, of course, define equivalent categories as categories between which there exists at least one isofunctor.

Free objects following an isofunctor

If \mathcal C and \mathcal C' are equivalent categories and \mathcal F is an isofunctor from \mathcal C to \mathcal C' with \mathcal F' its inverse isofunctor from \mathcal C' to \mathcal C, then we have:

If A is an object of \mathcal C and A' = \mathcal F(A), then (A', \iota_A) (with \iota_A the identity morphism on A) is a free \mathcal C' object on A following the functor \mathcal F'.

The proof is easy. If B' is any object of \mathcal C' and \beta a \mathcal C-morphism A \to \mathcal F'(B'), then for a \mathcal C'-morphism \gamma: A' \to B' to be such that \mathcal F'(\gamma) \circ \iota_A = \beta, necessarily \mathcal F'(\gamma) = \beta, that is (applying \mathcal F to each side) \gamma = \mathcal F(\beta). We can check that this \gamma = \mathcal F(\beta) indeed satisfies \mathcal F'(\gamma) \circ \iota_A = \beta. Hence there is one and only one \mathcal C'-morphism \gamma: A' \to B' such that \mathcal F'(\gamma) \circ \iota_A = \beta. This being the case for any object B' of \mathcal C' and any \beta \mathcal C-morphism A \to \mathcal F'(B'), the pair (A', \iota_A) is a free \mathcal C' object on A following \mathcal F'.

The “free object functor” in this case thus simply maps objects of \mathcal C to the same as \mathcal F does. It is easy to check that the effect of this free object functor on morphisms of \mathcal C is too just the same as that of \mathcal F. Thus, the free object functor in the case of equivalent categories following an isomorphism between them is just the inverse of that isomorphism.