Distributive property proof.

Question

I was reading this question here: Distributivity of categorical product and sum but I could not understand the statement of the OP that said "If $\textbf{C}$ is $\textbf{Set}$ or $\textbf{Top}$ I'm pretty sure the following "distributive law" holds: " so I was trying to prove the following generalized property:

$$\big(\amalg_{\alpha}A_{\alpha}\big)\times B \cong \amalg_{\alpha}(A_{\alpha} \times B),$$

where $A_{\alpha}$ and $B$ are topological spaces and $\cong$ means homeomorphic.

My thoughts are:

We are proving distributive property of categorical products(limits) over coproducts(colimits) in case of topological spaces and we know that the colimit topology condition for CW complexes can be stated as a subset $A$ of a topological space $X$ is open iff $A \cap X^{(n)}$ is open in $X^{(n)}$ for every $n.$ Where $X = \cup_{n=1} X^{(n)}$

But then how can I prove this for topological spaces? What is the limit topology then? how can I prove that the product topology can be distributed over the coproduct topology?

Any clarification will be greatly appreciated!

Edit:

Can we use the universal property of final and initial topology to prove this?

Answer 1

Unfortunately, category theory doesn’t help here, since this isn’t true for general (co)products. For instance, in the category of finite abelian groups, the product and coproduct are both the direct sum, so for $A_1=A_2=B=G$ this would be $(G\oplus G)\oplus G\cong(G\oplus G)\oplus(G\oplus G)$, which is obviously not true in general.

It’s true for sets, though, and usually things work out for topological spaces if you just do with the underlying sets what you would have done with sets and then check that the topologies come out right.

I’ll be identifying the components of a disjoint union with their images in the disjoint union in the usual way.

For sets, on the left we have elements $(a,b)$ with $a\in A_\alpha$ for some $\alpha$ and $b\in B$. On the right, we have elements of some $A_\alpha\times B$, which are also of the form $(a,b)$ with $a\in A_\alpha$ and $b\in B$. So the isomorphism of sets is pretty straightforward.

To check that the topologies match, you can use How to show 2 bases generate the same topology?, namely: Two bases $\mathcal B$ and $\mathcal B'$ generate the same topology iff for every $x\in U\in\mathcal B$ there is $U'\in\mathcal B'$ with $x\in U'\subseteq U$. (Here we’re actually only dealing with isomorphic topologies, not identical topologies, but the idea is the same.)

Given bases $\mathcal B$ and $\mathcal B'$ of $X$ and $X'$, respectively, a basis of $X\times X'$ is given by all Cartesian products $U\times U'$ with $U\in\mathcal B$ and $U'\in\mathcal B'$.

Given bases $\mathcal B_\alpha$ for $X_\alpha$, a basis of $\amalg_\alpha X_\alpha$ is given by $\amalg_\alpha\mathcal B_\alpha$.

Thus, if $\mathcal A_\alpha$ is a basis of $A_\alpha$ for all $\alpha$ and $\mathcal B$ is a basis of $B$, on both sides the basis elements have the form $U_\alpha\times U$ with $U_\alpha\in\mathcal A_\alpha$ and $U\in\mathcal B$, so they’re in one-to-one correspondence and generate isomorphic topologies.