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.
 A: 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.
