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Let $f$ be a continuous map from $X$ to $Y$ (with respect to their topologies). Let the equivalence relation ~ on $X$ be defined by: $x \sim y \iff f(x)=f(y)$ Then, $X/\sim$ is homeomorphic to $im(f)$ (equipped with the subspace topology).

A counterexample for the general case would be illuminating as well.

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  • $\begingroup$ Possible duplicate of Analogues of the Fundamental Isomorphism Theorems in Topology $\endgroup$ – Joseph Martin Mar 8 at 14:13
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    $\begingroup$ As stated, there could be a "coincidental" homeomorphism between $X /\!\sim$ and $im(f)$ that has nothing to do with the map $f$ itself. A better question would be to ask when the map $F : X/\!\sim \, \to im(f)$ that is induced by $f$ is a homeomorphism, namely the map $F([x])=f(x)$ where $[x]$ is the equivalence class of $x$; this holds if and only if $f$ is a quotient map (almost by definition). Is that what you intended? $\endgroup$ – Lee Mosher Mar 8 at 14:33
  • $\begingroup$ Yeah, this is what I intended! $\endgroup$ – Owen Tanner Mar 8 at 18:46
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Take any set $M$ and any two topologies $\mathfrak{T}_1, \mathfrak{T}_2$ on $M$ such that $\mathfrak{T}_1$ is strictly finer than $\mathfrak{T}_2$. Let $X_i = (M, \mathfrak{T}_i)$. Then the identity $id : X_1 \to X_2$ is continuous, but not a homeomorphism. Obviously the quotient map $p : X_1 \to X_1/\sim$ is a homeomorphism and we have $im(id) = X_2$. Thus the canonical map $id' : X_1/\sim \phantom{.} \to im(id)$ is not a homeomorphism (not even if both topolopies are Hausdorff).

I doubt that you will find "minimal" conditions assuring that the induced $f' : X/\sim \phantom{.} \to im(f)$ is a homeomorphism. In fact, the above example shows that not even for continuos bijections we have convincing minimal conditions.

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See this answer. Qiaochu Yuan states that this does not hold in general, but it does hold when both $X$ and $Y$ are Hausdorff.

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  • $\begingroup$ I have seen that answer, which indeed gets me close to the answer. But this is not minimal, is it? We can show, for example, a continuous bijection from a compact space onto a Hausdorff space must be a homeomorphism, so we can have just $X$ as a compact space and $Y$ as a Hausdorff, and this is sufficient. $\endgroup$ – Owen Tanner Mar 8 at 17:54
  • $\begingroup$ In Qiaochu Yuan 's answer $X,Y$ are required to be compact Hausdorff. $\endgroup$ – Paul Frost Mar 25 at 13:48

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