A proof with quotient topologies

Question

How would you prove that if $Q$ is the quotient of a topological space $X$ by and equivalence relation $\sim$, with the quotient topology. $f:X \to Y$ is continuous and has the property that whenever $x_1 \sim x_2$ then $f(x_1) = f(x_2)$. Then the map $\bar f :Q \to Y$ defined by $\bar f ([x]) = f(x)$ is continuous?

Answer 1

Just follow the definitions. Let $U$ be an open set in $Y$; you need to show that $\bar f^{-1}[U]$ is open in $Q$. By definition $\bar f^{-1}[U]$ is open in $Q$ if and only if

$$q^{-1}\left[\bar f^{-1}[U]\right]$$

is open in $X$, where $q:X\to Q$ is the quotient map. For any $A\subseteq Q$, $$q^{-1}[A]=\big\{x\in X:[x]\in A\big\}\;,$$

so you need to show that $$\left\{x\in X:[x]\in\bar f^{-1}[U]\right\}$$

is open in $X$. But $[x]\in\bar f^{-1}[U]$ if and only if $\bar f\big([x]\big)\in U$, and by hypothesis $\bar f\big([x]\big)=f(x)$, so

$$\left\{x\in X:[x]\in\bar f^{-1}[U]\right\}=\{x\in X:f(x)\in U\}=f^{-1}[U]\;.$$

Finally, $f$ is continuous, so $f^{-1}[U]$ is open in $X$, and we’re done.

Answer 2

Hint: If $U\subset Y$ is open, then show $\bar f^{-1}(U)$ is open in $Q$ by using that $f^{-1}(U)$ is open in $X$.

First, there is a natural map $\pi:X\to Q$, and the topology on $Q$ is defined so that $V\subset Q$ is open if and only if $\pi^{-1}(U)$ is open.

Now, as functions of sets, if $f:X\to Y$ has the property that $f(x)=f(x')$ when $x\sim x'$, then $f$ factors as $f=\bar{f}\circ \pi$. Then if $U$ is any subset of $Y$, then $f^{-1}(U)=\pi^{-1}(\bar f^{-1}(U))$.

If $f$ is continuous and $U\subset Y$ is open, then $f^{-1}(U)$ is open in $X$. But $f^{-1}(U)=\pi^{-1}(\bar f^{-1}(U))$. Letting $V=\bar f^{-1}(U)$, we see that $\pi^{-1}(V)$ is open in $X$, which means that $V$ is open in $Q$ by the definition of the quotient topology.

This means that $\bar f^{-1}(U)$ is open in $Q$ when $U$ is open in $Y$, that is $\bar f$ is continuous.