# How to show homeomorphism between a quotient subspace and the image of the subspace

I'm trying to show that if $$Y \subset X$$ and $$\sim$$ is an equivalence relation on X, Y an open set, and $$\pi:X\to X/\sim$$ is an open map, then $$Y/\sim$$ is homeomorphic to $$\pi(Y)$$ ($$\pi(Y)$$ will have relative topology of $$X/\sim$$).
I know that to be homeomorphic there has to exist a bijective function $$f:Y/\sim \to \pi(Y)$$ that is continuous and has a continuous inverse.
I think that using the identity map might work but I'm not sure how to show it.

This is not true. The map you seem to suggest is continuous and bijective, but may have a discontinuous inverse.

Let's start by fixing some notation, since there's a lot of very similar operations happening. Let's say that for $$x\in X$$ the set $$\langle x\rangle_X \in X/\sim$$ is the equivalence class of $$x$$ under $$\sim$$ and for $$y\in Y$$ the set $$\langle y\rangle_Y \in Y/\sim$$ is equivalence class of $$y$$ under the restriction of $$\sim$$ to $$Y$$.

While it's maybe not best to call it an identity map, it seems like you want to look at the definition $$f(\langle y\rangle_Y) = \langle y\rangle_X$$ and see if it's a homeomorphism. You'd first want to check that it's well-defined, but that's just an exercise with equivalence relations.

We then should also name the inclusion $$i:Y\rightarrow X$$ and the quotient maps $$\pi_X:X\rightarrow X/\sim$$ and $$\pi_Y:Y\rightarrow Y/\sim$$. It is worth proving the lemma that $$f\circ \pi_Y=\pi_X\circ i$$, which basically expresses the way in which $$f$$ is like an inclusion map. Moreover a set $$V$$ in $$Y/\sim$$ is open if and only if $$\pi_Y^{-1}(V)$$ is open - note that we will not actually use this property of $$\pi_X$$ in the proof of continuity, which hints at a slightly more general fact.

Then, you can let $$U$$ be an open set in $$X/\sim$$ and let $$V=f^{-1}(U)$$. Since $$U$$ is open, $$i^{-1}(\pi_X^{-1}(U))$$ is open as these maps are continuous. However, $$V$$ is open if and only if $$\pi_Y^{-1}(f^{-1}(U))$$ is open, but that set equals $$i^{-1}(\pi_X^{-1}(U))$$, which we just said was open, so $$f$$ is continuous.

The inverse of $$f$$ is given by $$f^{-1}(S) = S\cap Y$$ where $$S$$ is an equivalence class in $$X/\sim$$ and we note that $$S\cap Y$$ will be an equivalence class of $$Y/\sim$$ whenever it is non-empty - which is exactly when it is in the image of $$\pi_X(Y)$$. However, if you try to show this is continuous, you'll see that knowing that $$\pi_Y^{-1}(U)$$ is open doesn't really help show that $$f(U)$$ is open in $$\pi_X(Y)$$ - which is what you'd need to do.

And, indeed, this map needn't be continuous: it's possible for $$Y$$ to be unable to see how the equivalence relation sticks points together. For instance, take $$X=\mathbb R$$ and $$Y=\sqrt{2}\mathbb Z$$ and define $$x\sim y$$ to mean that $$x-y\in\mathbb Z$$. Then $$X/\sim$$ is a circle and $$Y/\sim$$ is just $$Y$$, since $$\sim$$ is the trivial relation on $$Y$$. However, the countable discrete space is not a subspace of the circle, so we see that the inverse function from $$\pi_X(Y)$$ to $$Y$$ is not continuous.