This problem appeared to me at first when studying topology of CW-complexes. I also needed ths fact when trying to prove that some graphs are contractible, when seen as complexes. Basically, I need to prove that, if $X$ has the weak topology with respect to some subspaces $A_{i}$, and $\mathbb{I}$ is the unit interval $[0,1]$, then the product $X\times \mathbb{I}$ has too the weak topology with respect to the spaces $A_{i}\times [0,1]$.
What I mean in ''$X$ has the weak topology with respect to the subspaces $A_{i}$'' is that for every subset $B\subseteq X$, $B$ is open if and only if $B\cap A_{i}$ is an open set in $A_{i}$ for every $i$. (this definition appears in the book "Topology", by J. Dugundji, page 131).
What I wanted to prove is the following:
Let $X$ be a topological space and $\{A_{i}\}_{i\in I}$ a family of subspaces whose union is $X$. If $X$ has the weak topology with respect to $\{A_{i}\}_{i\in I}$, then $X \times \mathbb{I}$ has the weak topology with respect to $\{A_{i}\times \mathbb{I}\}_{i\in I}$.
At first, I tried to prove it by definition: the problem is that while I can prove that the product topology is contained in the weak topology (by means of basic open sets), the converse didn't seem as easy, since I don't have a clear basis for $X\times \mathbb{I}$ with the weak topology. I have tried now a quotient topology argument, which goes as follows:
Firstly, consider the mapping $p\colon\bigsqcup_{i\in I}A_{i}\to X$ (where the domain is the disjoint union of the $A_{i}$'s) generated by the inclusions $\iota_{i}\colon A_{i}\to X$. It is clear from the definitions that $X$ has the weak topology with respect to $\{A_{i}\}_{i\in I}$ if and only if $p$ is a quotient map.
Secondly, by repeating the argument in the case of $X\times \mathbb{I}$, we have that $X\times \mathbb{I}$ has the weak topology with respect to the family $\{A_{i}\times \mathbb{I}\}_{i\in I}$ if and only if the map $q\colon\bigsqcup_{i\in I}(A_{i}\times \mathbb{I})\to X\times \mathbb{I}$ induced by the inclusions $\iota_{i}\times \operatorname{Id}_{\mathbb{I}}\colon A_{i}\times \mathbb{I}\to X\times \mathbb{I}$ is a quotient map. That said, I expressed $q$ as the composition of $\alpha \colon\bigsqcup_{i\in I}(A_{i}\times \mathbb{I})\to \big(\bigsqcup_{i\in I}A_{i}\big)\times \mathbb{I}$, which is the ''identity map": for every $(a,t)\in A_{i}\times \mathbb{I}$, $\alpha(a,t)=(a,t)$, where I consider that $a$ lies precisely in $A_{i}$, and $p\times \operatorname{Id}_{\mathbb{I}}\colon \big(\bigsqcup_{i\in I}A_{i}\big)\times \mathbb{I}\to X\times \mathbb{I}$. On one hand, $\alpha$ is a homeomorphism, since it is open, continuous and bijective. On the other hand, Whitehead's Theorem asserts that since $p$ is a quotient map, and $\mathbb{I}$ is a locally compact space, then $p\times \operatorname{Id}_{\mathbb{I}}$ is a quotient map, hence $q=(p\times \operatorname{Id}_{\mathbb{I}})\circ \alpha$ is a quotient map as a composition of quotient maps.
Finally, since $q$ is a quotient map, it follows that $X\times \mathbb{I}$ has the weak topology with respect to $\{A_{i}\times \mathbb{I}\}_{i\in I}$, which would make the proof complete.
Is this argument correct? And would it still be applicable when we change $\mathbb{I}$ for, say, a locally compact space?
Edit: I'll add my attempt at proving that $\alpha$ is a homeomorphism for completeness' sake:
$\bullet$ $\alpha$ is clearly bijective by construction.
$\bullet$ $\alpha$ is continuous: given a basic open set $B\times C\subseteq \big(\bigsqcup_{i\in I}A_{i}\big)\times \mathbb{I}$, we have that $B\cap A_{i}$ is open in $A_{i}$ for all $i$, so that $B\times C$ is also open in $A{i}\times \mathbb{I}$ for every $i$. Therefore, $B\times C=\alpha^{-1}(B\times C)$ is open in $\bigsqcup_{i\in I}(A_{i}\times \mathbb{I})$.
$\bullet$ $\alpha$ is open: it is enough (since $\bigsqcup_{i\in I}A_{i}$ has the disjoint union topology) to prove that a set of the form $B\times C$, where $B\subseteq A_{i}$ (for some $i$) is open and $C\subseteq \mathbb{I}$ is open has an open image. But $\alpha(B\times C)=B\times C$ is open, since $B\cap A_{j}=\emptyset$ if $j\neq i$, and $B\cap A_{i}=B$, which are all open sets. Therefore, $\alpha(B\times C)$ is open as a product of open sets in their respective factors.
Thank you in advance!
Yes, the argument is essentially correct for any locally compact space $\mathbb{I}$. Good job! To make the argument rigorous, we need to replace $A_i$’s in the disjoints sums by its pairwise disjoint homeomorphic copies.