# Inverse of map

1. Apr 4, 2012

consider we have a map.
what condition should have our map that it has inverse?

2. Apr 4, 2012

### morphism

Since this is in the topology & geometry forum, I assume by "map" you don't just mean a set-theoretic map between two sets, but maybe a continuous map between topological spaces, or a smooth map between smooth manifolds, or something else of this sort. And I interpret your question as asking "when is the inverse of such a map also a map of the same type (e.g. inverse of continuous map is continuous, inverse of smooth is smooth)?" Of course to even speak of the inverse we need to know that our map is a bijection, so at least there's this obvious necessary condition. However, there is no "useful" general answer to this interpretation of the question. E.g. there are no easily-checked criteria for determining when a continuous bijection between two general topological spaces has a continuous inverse. You could say something like "it has a continuous inverse iff it's an open map", but that's just superficially different way of saying "it has a continuous inverse iff it has a continuous inverse".

Did I understand your question correctly or did you have something else in mind?

3. Apr 4, 2012

### Bacle2

Another approach:

Actually, there is a nice little rule that helps some times to tell when a continuous bijection is a homeomorphism: A continuous bijection f:X-->Y , where X is compact and Y is Hausdorff, is a hemeomorphism, but that still leaves a lot of maps out.

Maybe another issue is to check for known topological invariants: if X (equiv. Y) has this invariant and Y(equiv. X) does not, and f:X-->Y is a bijection, then f^{-1} is not continuous.