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.
On
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.