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I got a question on my topology exam that I did not manage to solve. The exam is over, so now I ask here.

Let $X$ be a topological space. For each equivelance relation $R$ on $X$, let $X/R$ be the set of $R$-equivalence classes in $X$, and let $q_R: X \to X/R$ be the canonical quotient map. We say that $R$ is good if the quotient space $X/R$ is Hausdorff.

Define the product space $$Y = \prod_{\text{R is good}} X/R.$$

Let $f:X \to Y$ be the function with $\pi_R \circ f = q_R$ for each good $R$, where $\pi_R: Y \to X/R$ is the $R$-th projection mapping. Let $Z = f(X)$ be its image, with the subspace topology from $Y$.

Define $g:X \to Z$ to be the corestricted function $f$, so that $g(x) = f(x)$ for each $x \in X$.

At last, the problem: prove that $g:X \to Z$ is a quotient map.

What I have done already is that I have shown that $g$ is a continuous surjection (with Hausdorff image Z. I'm not sure if this is relevant, but one of the earlier sub-problems asked me to do so.) This part was not too challenging.

So I have shown the "$\implies$" direction of the required statement $$U \in Z \text{ is open} \iff g^{-1}(U) \in X \text{ is open}. $$ But the other direction seems pretty tricky with no knowledge of $X$.

I've tried showing that $g$ is an open and a closed map, but I can't determine if this is true. Working with the product topology of an arbitrary cardinality does my head in!

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$f$ is continuous by the universal property of initial topologies. The corestricted $g$ is then also continuous. By definition it's onto and $Z$ is Hausdorff as a subspace of the Hausdorff $Y$ (as all $R$ are good, all factor spaces are by definition Hausdorff, and arbitrary products of Hausdorff spaces are Hausdorff, and subspaces inherit it too).

If $\pi_R^{-1}[U]$ is a subbasic shaped subset of $Y$ then $f^{-1}[\pi_R^{-1}[U]] = (\pi_R \circ f)^{-1}[U]= q_R^{-1}[U]$ is open iff $U$ is open in $X{/}R$. Try to use that to show $g$ quotient.

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