# Homeomorphism between $X/\tilde~$ and $Y/\tilde~'$

Let $X,Y$ be topological spaces, $\tilde~$ equivalence relation on space $X$ and $\tilde~'$ equivalence relation on space $Y$.

If $f:X\rightarrow Y$ is homeomorphism, where $x_{1}\tilde~ x_{2}\Leftrightarrow f(x_{1})\tilde~' f(x_{2})$, so $f$ induces a homeomorphism $X/\tilde~\rightarrow Y/\tilde~'$.

I am using the theorem that says that bijection $f:X\rightarrow Y$ is a homeomorphism, iff $U\subseteq X$ open iff $f(U)\subseteq Y$ open.

Now, that part where I must show the equivalence mentioned above is clear.

But my problem is the bijectivity of $\overline{f}:X/\tilde~\rightarrow Y/\tilde~'$.

The condition between the equivalence relations means that for every point in any equivalence class it's image is in the image of it's equivalence class. That and the fact $f$ is homeomorphism should give me enough information to say that $\overline{f}$ is indeed bijection. But I am not sure.

Thoughts, ideas?

$\def\si{\mathord\sim}$To see that $\bar f$ is one-to-one, suppose $\bar f(x_1/\si) = \bar f(x_2/\si)$, that is $f(x_1) = f(x_2)$ by definition of $\bar f$, therefore $f(x_1) \sim' f(x_2)$, hence $x_1 \sim x_2$. So we have $x_1/\si = x_2/\si$.
To see that $\bar f$ is onto, suppose $y/\si' \in Y/\si'$, as $f$ is onto, there is $x \in X$ with $f(x) = y$, but then $\bar f(x/\si) = y/\si'$.