# continuous function that factors over quotient topology

Given a metric space $$(Y,l)$$ and a continuous mapping $$f:X \mapsto Y$$ (w.r.t $$(X,\tau_d)$$ for some pseudometric $$d$$) and $$R \subset X\times X$$ with $$a \sim b$$ $$\Leftrightarrow$$ $$d(a,b)=0$$.

I'm now trying to show that there exists a continuous map $$g: X/R \mapsto Y$$ (w.r.t quotient topology) for which $$f=g\circ\pi_R$$ holds. ($$\pi_R$$ maps $$x$$ to its equivalence class)

At first i thought about defining $$g([x])=f(x)$$ but i am having some trouble showing it is well defined, given that we only know that $$Y$$ is a metric space.

Can someone help me out on how to define $$g$$?

It's been a little while since I've done point set topology, but I think this is correct: begin with $$d(x,x')=0\Rightarrow\forall\ U\in\tau_d\ s.t.\ U\ni x,x'\in U$$ That is, all open sets in $$X$$ containing $$x$$ also contain all points $$x'$$ with $$d(x,x')=0$$. Then note that, by positivity of $$l$$, $$\forall y\in Y,\ (\forall V\ni y,\ y'\in V) \Rightarrow y'=y.$$ Finally conclude that, for any $$x,x'\in X$$ with $$d(x,x')=0$$, $$\forall\ \text{open}\ V\ni f(x),\ f^{-1}(V)\in\tau_d\Rightarrow x'\in f^{-1}(V)\Rightarrow f(x')\in V$$ $$\therefore f(x')=f(x)$$

We've thus shown that $$d(x,x')=0\Rightarrow f(x)=f(x')$$, which makes your definition of $$g$$ well-defined. Your choice of $$g$$ is good otherwise.