olivierd – Precisely

The composition of polynomials

Definition

Let $\K$ be a commutative field. The polynomials $\K[X]$ form a commutative unitary algebra over $\K$ containing an element $X$ such that the following universal property holds:

For any unitary algebra $A$ (commutative or otherwise) and any element $a \in A$ , there exists exactly one unitary algebra morphism $\phi: \K[X] \to A$ such that $\phi(X) = a$ .

If $P$ and $Q$ are elements of $\K[X]$ , let $\gamma_Q: \K[X] \to \K[X]$ be the unique morphism such that $\gamma_Q(X) = Q$ . Then the composition $P(Q)$ of $P$ and $Q$ is defined by $P(Q) = \gamma_Q(P)$ .

Effect of composition of polynomials on the corresponding functions

In terms of polynomial functions, the composition of polynomials is the equivalent to the composition (via “ $\circ$ ”) of the associated functions.

Let us first define the function that is associated with a polynomial.

The set $\K^\K$ of functions $\K \to \K$ is made into a commutative and unitary associative algebra by the usual operations of summing ( $(f + g)(x) = f(x) + g(x)$ , multiplication by a scalar ( $(af)(x) = a(f(x))$ and internal multiplication ( $(fg)(x) = f(x) g(x)$ ).

Hence, following the universal property of the algebra of polynomials, there exists a unique morphism $\phi: \K[X] \to \K$ such that $\phi(X) = \Id_\K$ .

By definition, the function $p$ associated with a polynomial $P$ is simply $p = \phi(P)$ .

We wish to show that if $P$ and $Q$ are polynomials and $p$ and $q$ the respective associated functions, the function associated with $P(Q)$ is $p \circ q$ .

Let $Q$ be a polynomial and $q = \phi(Q)$ . Let us consider the mapping $\alpha: \K[X] \to \K^\K, P \mapsto \phi(P) \circ q$ .

It is easy to check that $\alpha$ is a morphism.

Furthermore, $\alpha(X) = \phi(X) \circ q = \Id_\K \circ q = q$ .

On the other hand, $\phi \circ \gamma_Q$ is also a morphism from $\K[X]$ to $\K^\K$ , and $(\phi \circ \gamma_Q)(X) = \phi(\gamma_Q(X)) = \phi(Q) = q$ .

Thus both $\alpha$ and $\phi \circ \gamma_Q$ are morphisms $\K[X] \to \K^\K$ and they agree on the value $X$ . The universal property of polynomials entails that they are equal.

Hence for any $P \in \K[X]$ , $(\phi \circ \gamma_Q)(P) = \alpha(P)$ .

That is, $\phi(P(Q)) = \phi(P) \circ \phi(Q)$ .

Associativity

If $P$ , $Q$ and $R$ are polynomials in $\K[X]$ , we wish to show that:

$P(Q(R)) = (P(Q))(R)$

Using the definition of composition, this translates into:

$\gamma_{\gamma_R(Q)}(P) = \gamma_R(\gamma_Q(P))$

That is:

$\gamma_{\gamma_R(Q)}(P) = (\gamma_R \circ \gamma_Q)(P)$

We need to show that for any $Q$ and $R$ in $\K[X]$ , the two $\K[X] \to \K[X]$ mappings $\gamma_{\gamma_R(Q)}$ and $\gamma_R \circ \gamma_Q$ are identical.

Both are endomorphisms of $\K[X]$ .

We have:

$\gamma_{\gamma_R(Q)}(X) = \gamma_R(Q)$ by definition of $\gamma_{\gamma_R(Q)}$ .
$(\gamma_R \circ \gamma_Q)(X) = \gamma_R(\gamma_Q)(X)) = \gamma_R(Q)$ by definition of $\gamma_Q)$ .

Hence both morphisms agree on the image of $X$ . By virtue of the universal property of polynomials, they are equal.

Hence for any $P \in \K[X]$ , we indeed have $\gamma_{\gamma_R(Q)}(P) = (\gamma_R \circ \gamma_Q)(P)$ , that is, $P(Q(R)) = (P(Q))(R)$ .

Effect of composition on the evaluation of polynomials

todo

Derivation of the composition of polynomials

todo

Equivalent alternative definitions concerning filters

Definition of a filter

Let $E$ be a set.

Traditional definition of a filter on $E$ : A collection $\mathcal F$ of subsets of $E$ is a filter on $E$ iff $\emptyset \notin E$ ; and $\forall A \in \mathcal F, \forall B \subseteq E, (B \supseteq A \implies B \in \mathcal F)$ ; and $\forall A, B \in \mathcal F, A \cap B \in \mathcal F$ .

Alternative definition: A collection $\mathcal F$ of subsets of $E$ is a filter on $E$ iff $\emptyset \notin E$ ; and $\forall A, B \subseteq E, ((A, B) \in \mathcal F^2 \iff A \cap B \in \mathcal F)$ .

The two definitions are equivalent.

Definition of the image of a filter

Let $E$ and $F$ be sets, $\mathcal F$ a filter on $E$ and $f: E \to F$ .

Traditional definition: The filter on $F$ image of $\mathcal F$ by $f$ is $(f^{\rightarrow \rightarrow}(\mathcal F))^{\uparrow F}$ .

Alternative definition: The filter on $F$ image of $\mathcal F$ by $f$ is $\{\ B \subseteq F\ |\ f^\leftarrow(B) \in \mathcal F\ \}$ .

Again, the two definitions are equivalent.

Margin between a compact subspace and an including open subspace

In Chapter 31 (“Compact-Open Topology”), the fact that, on the set of the continuous mappings $\R \to \R$ the second topology (the uniform convergence topology) is finer than the first (the compact-open topology) is said to be “clear” but actually needs some non-trivial proof. Part of this is the lemma that I state and prove below.

For any $x \in E$ and real $r > 0$ , let $\dot B(x, r)$ be the open ball centered on $x$ and with radius $r$ . We know that an open ball is an open subset.

The lemma says:

Let $(E, d)$ be a metric space, $O$ an open subspace of $E$ and $C \subseteq O$ a compact subspace of $E$ . Then there exists $r > 0$ such that for all $x$ in $C$ , $\dot B(x, r) \subseteq O$ .

Proof:

For each $x \in C$ , since $x \in O$ , there exists an $r_x > 0$ such that $\dot B(x, r_x) \subseteq O$ .

The collection of all $\dot B(x, \frac 1 2 {r_x})$ with $x \in C$ is an open cover of $C$ . Hence there exists a finite $J \subseteq C$ such that $\{\ \dot B(x, \frac 1 2 {r_x})\ \}_{x \in J}$ covers $C$ .

Since $J$ is finite, there exists a real $r > 0$ such that for all $x \in J$ , $r \le \frac 1 2 {r_x}$ .

Let $x$ be an eventual point of $C$ .

Since $\{\ \dot B(z, \frac 1 2 {r_z})\ \}_{z \in J}$ covers $C$ , there exists $z \in C$ such that $x \in \dot B(z, \frac 1 2 {r_z})$ , that is, $d(z, x) < \frac 1 2 {r_z}$ .

Let $y$ be an eventual point of $\dot B(x, r)$ .

We have $d(z, x) < \frac 1 2 {r_z}$ and $d(x, y) < r \le \frac 1 2 {r_z}$ , which leads to $d(z, y) < r_z$ , that is, $y \in \dot B(z, r_z)$ . But $z$ is in $C$ and $r_z$ was taken such that $\dot B(z, r_z) \subseteq O$ . Hence $y \in O$ .

Hence $\dot B(x, r) \subseteq O$ .

We thus have $r > 0$ such that $\forall x \in C,\ \dot B(x, r) \subseteq O$ . $\blacksquare$

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 either $V \cap K$ 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 both $a$ and $b$ , then you know $(a, b)$ .

Theorem:

For every $c_1, c_2 \in Z$ , 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 set (singleton sets 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 $\gamma_1$ and $\gamma_2$ as the constant mappings, both $Z' \to Z$ , 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.