Consider topological spaces $X,Y$ and a map $f : X \to Y$ (not necessarily continuous). We say $f$ reflects limits if for all nets $x : D \to X$ and $x_\infty \in X$ we have
$$f\circ x \to f(x_\infty) \Rightarrow x \to x_\infty.$$
It is somewhat like saying "if $f$ had an inverse, then that inverse would be continuous".
Is there a known characterization of this property in terms of well-known properties?
Special cases ($X,Y$ Hausdorff, metric spaces, TVS, Banach spaces ...) are also of interest.
Assume that $X$ is $T_1$. Note that I will weaken your assumptions a bit, because if $X$ is not Hausdorff then limits are not unique. So your definition is not clear anymore. But let's forget about that problem for the moment.
If $f$ is not injective then it cannot have this property. Indeed, assume that $f(x)=f(y)$ for some $x\neq y$ and $f$ has the desired property. Then you just take an infinite sequence
$$x_n=\begin{cases} x&\text{if }n\text{ is odd}\\ y&\text{if }n\text{ is even}\\ \end{cases}$$
Since $f(x_n)$ is constant then it is convergent. By the assumption that implies that $x_n$ is convergent. But since $X$ is $T_1$ then it is not. Contradiction. Note that here I don't care about (possibly non-unique) limits. Being convergent is enough to get this contradiction.
So it follows that the property implies that $f$ is injective. And if that's the case then $f^{-1}$ (which is defined on the image of $f$) has to be continuous (as you've noted yourself). And that's the full characterization I guess.
Also note that I only used sequences in the proof. Which is weaker then your assumptions.
For non-$T_1$ spaces you can find non-injective examples. Can't help you beyond that.