Theorems about the direct product of two topological spaces

olivierd

Jul 20, 2017 9 min read

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 $E$ . $\blacksquare$