I know that if $X/\sim$ is a quotient space and $f:X \mapsto Y$ is a continous map, then there exists a unique continous map $g:X/\sim \mapsto Y$ such that $$ g\circ q= f, $$ where $q$ is the universal quotient map.
My question is that, is such a statement also true for lower semi-continous functions?
Suppose that $\sim$ is an equivalence relation on $X$, and $f:X\to Y$ is a lower semi-continuous function that respects the equivalence relation $\sim$ on $X$. Recall that $f$ is lower semi-continuous if and only if the set $$U_a = \{x\in X\ |\ f(x)>a\}$$ is open in $X$ for all $a\in\mathbb{R}$. Then, as $f$ respects the equivalence relation on $X$, there is a map (of sets for now) $\bar{f}:(X/\sim) \to Y$ making the diagram: $$\require{AMScd} \begin{CD} X @>{\pi}>> X/\sim\\ @V{f}VV @VV{\bar{f}}V\\ Y @>>{id_Y}> Y \end{CD}$$ commute. The map $\bar{f}$ will be lower semi-continuous if we can show that the set $$\{[x]\in X/\sim\ |\ \bar{f}([x])>a\} = \{[x]\in X/\sim\ |\ f(x)>a)\} = \pi(U_a)$$ is open for all $a\in\mathbb{R}$. However, as $f$ is lower semi-continuous, $U_a$ is open, and as $\pi$ is an open map, $\pi(U_a)$ must be open as well. Since $a\in\mathbb{R}$ was arbitrary, we are done.
The same proof can be adapted for upper semi-continuous functions.