$(X\times [0,1])/\sim$ homeomorphic to $X\times S^1$

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The equivalence relation $\sim$ is given by $(x,0)\sim (x,1) \quad\forall x\in X$.

I already know that $[0, 1]/ \{0, 1\}$ is homeomorphic to $S^1$ but have problems showing this for the given product space as I know nothing about the topological space $X$ and therefore can't use e.g. compactness.

So far, I defined $$f:X\times [0,1]\to X\times S^1,\quad f(x,y):=(x,exp(2\pi iy))$$ Therefore I get the induced map $g:(X\times [0,1])/\sim \to X\times S^1$ and I know that $g$ is a continuous bijection. Now it remains to be shown that $g^{-1}$ is continuous, too, and that is what I'm struggling with. Is there any theorem I can use or do I write out $g^{-1}$ to show the continuity via open sets? If so, can someone give me a starting point? I still have problems understanding the product and quotient topology.

Thank you in advance!

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Hint: if you can show that $g$ is open or closed, then it follows that $g$ is a homeomorphism, because it’s already bijective. Do you see why it’s open? Use the product topology.

Edit: Note that $[0,\epsilon)\cup (1-\epsilon,1]$ is open in $[0,1]$ because its complement $[\epsilon,1-\epsilon]$ is closed, hence its image, a neighbourhood of $0 \sim 1$, is also open.