Quotient Space and Product

Question

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!

Answer 1

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

Theorem (Whitehead): Let $X,Y,Z$ be topological spaces with $Z$ locally compact. If $f:X\to Y$ is a quotient map and $id:Z\to Z$ is the identity then $f\times id:X\times Z\to Y\times Z$ is a quotient map.