Prove that the induced map $\tilde{f}: X/\sim \to Y/\square$ is continuous; moreover, if $f$ is an identification, then so is $\tilde{f}$. 
Let $X$ and $Y$ be spaces with equivalence relations $\sim$ and $\square$, respectively, and let $f: X \to Y$ be a continuous map preserving the relations (if $x \sim x'$, then $f(x) \square f(x')$). Prove that the induced map $\tilde{f}: X/\sim \to Y/\square$ is continuous; moreover, if $f$ is an identification, then so is $\tilde{f}$.

Let $U$ be open in $Y/\square$. Since we're dealing with the quotient spaces $X/\sim$ and $Y/\square$  they come equipped with the quotient maps $q :X \to X/\sim$ and $p : Y \to Y/\square$. Now if $\tilde f = p \circ f \circ q^{-1}$, then $$\tilde f ^{-1}(U)=(p\circ f \circ q^{-1})^{-1}(U)=q(f^{-1}(p^{-1}(U)))$$ and what I know is that $f^{-1}(p^{-1}(U))$ is open as $f$ and $p$ are both continuous maps. However this arises two questions

*

*Is the diagram resulting from $\tilde f = p \circ f \circ q^{-1}$ commutative? That is can I define the induced map like this?

*The continuity of $\tilde f$ depends on $p$ being an open map as $f ^{-1}(U)=(p\circ f \circ q^{-1})^{-1}(U)=q(f^{-1}(p^{-1}(U)))$ is only open if $p$ is an open map. So is it?

 A: First of all, $\tilde f$ is defined as follows: Given $x \in X$, $\tilde f$ maps the $\sim$-equivalence class of $x$ to the $\square$-equivalence class of $f(x)$. Note that the hypothesis
$$
\forall x_1,x_2 \in X : \quad x_1 \sim x_2 \implies f(x_1) \,\square\, f(x_2)
$$
guarantees that $\tilde f$ is well-defined.
Said in other way, if $\pi_\sim \colon X \to X/{\sim}$ and $\pi_\square \colon Y \to Y/\square$ are the functions which maps each element to its equivalence class, then $\tilde f$ is “defined” by $$\tilde f \circ \pi_\sim = \pi_\square \circ f.$$
Now, I claim that this problem is a particular application of the following:

Proposition. Let $q \colon X \to Q$ be a quotient map$^*$, and $g \colon X \to Y$ be a continuous map satisfying:
$$
\forall x_1,x_2 \in X : \quad q(x_1) = q(x_2) \implies g(x_1) = g(x_2).
$$
Then the map $\bar g \colon Q \to Y$ “defined” by $\bar g \circ q = g$ is continuous.

Notice that if $Q=X/{\sim}$, $q=\pi_\sim$ and $g = \pi_\square \circ f$, then $\bar g$ is precisely $\tilde f$ and we’re done.
Proof of the proposition: Take an open set $V$ of $Y$. As $g$ is continuous, $g^{-1}[V]$ is open in $X$. But $g^{-1}[V] = (\bar g \circ q)^{-1}[V] = q^{-1}[\bar g^{-1}[V]]$. Hence, $\bar g^{-1}[V]$ is a subset of $Q$ such that its inverse image under $q$ is open in $X$; therefore, $\bar g^{-1}[V]$ is open in $Q$. $\blacksquare$

$^*$Also known as an identification; that is, $q$ is a surjective continuous map with the following property: If $U$ is a subset of $Q$ such that $q^{-1}[U]$ is open in $X$, then $U$ is open in $Q$.
