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

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