Theorems about the direct sum of two topological spaces

olivierd

Jul 19, 2017 4 min read

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