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Let $X$ be a compact topological space and let $\sim$ be an equivalence relation on $X$ with the property that $X/{\sim}$ is Hausdorff. Let $\sim'$ be the equivalence relation on $X\times [0,1]$ defined by $(x,t)\sim'(x_1,t_1)\Leftrightarrow x\sim x_1$ and $t= t_1.$ Prove that $(X\times [0,1])/{\sim'}$ is homeomorphic to $(X/\sim)\times[0,1].$

My thoughts:

Consider the map $$\begin{align} f\colon X\times[0,1]&\to (X/{\sim})\times[0,1]\\ (x,t)&\mapsto(\bar{x},t) \end{align}$$

Then it is clear that $f$ is a well-defined surjective and continuous map. Moreover, since $X$ and $[0,1]$ are compact, $X\times [0,1]$ is compact. If we can say that $X\times [0,1]$ is Hausdorff, then $f$ is a closed map. Then we would have that $f$ is actually a quotient map and there is an induced homeomorphism $$\begin{align} (X\times[0,1])/{\sim'}&\to (X/{\sim})\times[0,1]. \end{align}$$

But can we say that $X$ is Hausdorff if $X/{\sim}$ is?

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  • $\begingroup$ (You can't say that $X$ is Hausdorff if $X/\sim$ is.) $\endgroup$ – Thomas Andrews Dec 30 '16 at 16:53
  • $\begingroup$ Yes I do mean $t=t_1$ $\endgroup$ – user346096 Dec 30 '16 at 16:53
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What you have is that there is a continuous one-to-one and onto function $X\times [0,1]/{\sim'}\to (X/{\sim})\times[0,1]$. Not all continuous one-to-one and onto functions are homeomorphisms, but there is one theorem:

A continuous one-to-one and onto map from a compact space to a Hausdorff space is a homeomorphism.

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You don't need $X\times [0,1]$ to be Hausdorff.

Since it is compact, every closed subset is also compact, and so is its image by $f$. But since $(X/{\sim})\times[0,1]$ is Hausdorff, its image is closed, so $f$ is a closed map, and -- since it's surjective -- a quotient map.

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