One of the issues any beginning student of topology has the square up against is that in general you can't break a space up into smaller subspaces and expect that the original space will be the coproduct of the subspaces. For instance, the interval $[0,1]$ is the set-theoretic disjoint union of $[0,1/2)$ and $[1/2,1]$, however if we take a coproduct in the category of topological spaces, the canonical map $[0,1/2) \sqcup [1/2,1] \rightarrow [0,1]$ fails to be a homeomorpism. To see this, consider the inverse function, and notice that it's discontinuous; for example, the sequence $(0,1/4,3/8,\ldots,1/2-(1/2)^n,\ldots)$ converges to $1/2$ in $[0,1]$ but fails to converge in the coproduct space.
It seems to be possible to avoid this problem as follows:
- Instead of partitioning your space, you cover it with closed subspaces
- Instead of taking a coproduct, you take a colimit of an appropriate diagram.
For example, if $\{A,B\}$ is a closed cover of a space $X$, then it seems to be the case that the pushout of $$A \cap B \hookrightarrow A, \;A \cap B \hookrightarrow B$$ is canonically isomorphic to $X$. For example, if $A=[0,1/2]$ and $B=[1/2,2],$ then the pushout of $A$ and $B$ over the point they share in common seems to be $[0,1]$.
Furthermore, it seems likely that there's a version of this for an finite closed cover of a space $X$, in which one takes a colimit of a diagram containing an object for each pairwise intersection and a pair of inclusions for each such pairwise intersection.
(I don't think this really works for infinite closed covers; and $T_1$ space has its singletons as a closed cover, after all.)
Question. Is this correct, and if so, where can I learn more (e.g. standard notation, proof of correctness, etc.) If not, what would be a counterexample?