What our the minimal requirements on the topologies on $X$ and $Y$ such that the following statement is true?

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.

• Possible duplicate of Analogues of the Fundamental Isomorphism Theorems in Topology – Joseph Martin Mar 8 at 14:13
• 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? – Lee Mosher Mar 8 at 14:33
• Yeah, this is what I intended! – Owen Tanner Mar 8 at 18:46

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.
See this answer. Qiaochu Yuan states that this does not hold in general, but it does hold when both $$X$$ and $$Y$$ are Hausdorff.
• 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. – Owen Tanner Mar 8 at 17:54
• In Qiaochu Yuan 's answer $X,Y$ are required to be compact Hausdorff. – Paul Frost Mar 25 at 13:48