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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 $g$?

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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.

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