It is known that product topology and quotient space behave not as well as we wish. Nonetheless while I was reading some other posts on MSE, it seems that in some special cases the following claim does hold:
Claim: Let X be a topological space, and $\sim $ an equivalence relation on $X$. Then $(X\times I)/\sim'$ is homeomorphic to the space $(X/\sim)\times I$.
Here $I$ is the interval $[0,1]$ in $\mathbb {R}$ and $\sim'$ is the equivalence relation on $(X\times I)$ defined by $(x,t)\sim' (x',t')$ if $x\sim x'$ and $t=t'$.
I can see that the continuous map $p\times \textrm{id}:X\times I\rightarrow (X/\sim) \times I$ induces a continuous bijection from $(X\times I)/\sim'$ to $(X/\sim)\times I$ (,which is true in more general case). According to what I have found so far it seems that the (local) compactness of $I$ is essential in proving that the inverse of the induced map is continuous, but I am stuck here. So my question is
How can one prove the above claim? (And if local compactness is used in the proof, how is it used?)
Thanks in advance!
Best Answer
So you have
$$f:(X\times I)/\sim'\to (X/\sim)\times I$$ $$f([x,t])=([x],t)$$
and (as you've noted) this is a well defined continuous bijection. We can easily find the inverse:
$$g:(X/\sim)\times I\to (X\times I)/\sim'$$ $$g([x], t)=[x,t]$$
and the question is whether $g$ is continuous. So pick an open subset $U\subseteq (X\times I)/\sim'$ and let $\pi:X\times I\to (X\times I)/\sim'$ be the projection. Let $V:=\pi^{-1}(U)$. It is obviously open.
Now let $p:X\to X/\sim$ be the other projection and let $\tau:X\times I\to (X/\sim)\times I$ be given by $\tau(x,t)=(p(x), t)$. In other words $\tau=p\times id$ in your notation. Note that $\tau^{-1}(g^{-1}(U))=V$. Now if we knew that $\tau$ is a quotient map then we are done because that would imply that $g^{-1}(U)$ is open.
So when is $\tau$ a quotient map? For that we need $I$ to be locally compact and this is known as