This is a practice final exam.
My questions: 1) When the question defines X/~ as quotient topology. Does that mean I can write: $q: X \rightarrow X/\sim $ 2) Specifically, I am trying to prove that $\bar f$ is continuous. and $\bar f$ is bijective.
Note: I have already proven that inverse of $\bar f$ is continuous.
On
First check (this is easy) that $\sim$ defined by $x \sim y$ iff $f(x) = f(y)$ is an equivalence relation.
So we have the set of classes $X/\sim = \{[x]: x \in X\}$ and a standard quotient map $q: X \rightarrow X/\sim$ with $q(x) = [x]$. By definition $X/\sim$ is made into a topological space by giving it the quotient topology with respect to $q$. I.e. a set $U \subseteq X/\sim$ is open exactly when $q^{-1}[U] = \{x \in X: [x] \in U \}$ is open.
Now, $\sim$ is defined by using $f$. This means that if we have any class $[x] \in X/\sim$, we can pick any point $p$ in it, and have the same value $f(p)$ regardless of the $p$ we pick, so we could pick the representative $x$ of $[x]$ and define $\tilde{f}([x]) = f(x) \in Y$. So this defines a function from the set of classes of $X$ (under $\sim$) to $Y$. And this is well-defined (if $[x] = [x']$ then this means that $x \sim x'$ and so $f(x) = f(x')$), and we get the same value.
Both $X/\sim$ and $Y$ have topologies so we can discuss continuity. If $U$ is open in $Y$, then is $\tilde{f}^{-1}[U]$ open? By the definition of the quotient topology on $X/\sim$ this is exactly the case when $q^{-1}[\tilde{f}^{-1}[U]]$ is open in $X$. And $x \in q^{-1}[\tilde{f}^{-1}[U]]$ iff $q(x) = [x] \in \tilde{f}^{-1}[U]$ iff $\tilde{f}([x]) = f(x) \in U$ iff $x \in f^{-1}[U]$. So
$$q^{-1}[\tilde{f}^{-1}[U]] = f^{-1}[U]$$
and so this is indeed open by continuity of $f$.
$\tilde{f}$ is surjective as $f$ is surjective (check this).
$\tilde{f}$ is injective because we precisely identify values of $f$ and no more: suppose $\tilde{f}([x]) = \tilde{f}([x'])$. This means by definition of $\tilde{f}$ that $f(x) = f(x')$ and so $x \sim x'$ and thus $[x] = [x']$. So we have injectivity.
All we have used so far for $f$ is continuity and surjectivity.
If $f$ is a quotient map, then we note that for $O \subseteq X/\sim$ open, we have that $f^{-1}[\tilde{f}[O]] = \{x \in X: f(x) \in \tilde{f}[O]\}$ just equals $q^{-1}[O]$ (check this!) and so is open by continuity of $q$, and as $f$ is a quotient map, $\tilde{f}[O]$ is then open in $Y$. This makes $\tilde{f}$ an open, continuous, bijection so a homeomorphism between $X /\sim$ and $Y$.
Your fact is a special case, as a continuous surjection from a compact space to a Hausdorff space is closed, and so in particular a quotient map.
Presumably you want $\bar{f}: (X/\sim) \to Y$ to be your homeomorphism, correct? One way to define $\bar{f}$ is to let $\bar{f}([p]) = f(p)$ where $p \in X$ but $[p] \in X/\sim$. Now if $U \subseteq Y$ then to prove $\bar{f}$ continuous we want $\bar{f}^{-1}(U)$ to be open in $X/\sim$. This set is equivalent to writing $(q\circ f^{-1})(U)$ and this is open since both $f$ and $q$ are continuous.