Consider an “analytically nice” bijective function ; the question I want to explore is, when is its inverse also analytically nice?

Let’s start in the topological category. Let be a continuous, bijective map of topological spaces. When is a homeomorphism? The inverse in general need not be continuous:

Let be set with more than one element. Let denote with the discrete topology: every subset of is open. Let denote with the indiscrete topology: the only open sets of are the empty set and all of itself. The identity function is clearly continuous. This is equivalent to saying that the discrete topology on is finer than the indiscrete topology. However, the converse is false: is not continuous, for example, the preimage of a singleton set is not open. Let be given by The function is continuous by construction. However, the inverse is not continuous: is not open.

Let’s return to the general case . The inverse is continuous iff, for every open, is open in . Thus, is continuous iff is an open map.

Now, let’s turn to real functions. Let be a continuous one-to-one function, that is, is a bijection with its image.

Proposition 1 The inverse is continuous.

Proof: It is a straightforward consequence of the intermediate value theorem that is either (strictly) decreasing or increasing. Suppose is increasing; otherwise replace by . Since is increasing, so is . Let and set . Let be given. By increasing, . Let be the minimum of and . Suppose . Then . Appying , we get

as desired.

(What about continuity at a point?)

Now, what about differentiability? Again, let is a one-to-one, continuous function and suppose is differentiable at . Set . Suppose were differentiable at . Since , by the chain rule, we have

Thus, if , cannot be differentiable at . For example, consider and . In fact, this is the only obstruction to being differentiable.

Proposition 2 If , then is differentiable at , and, moreover,

Proof: For any sufficiently small, there exists a unique such that . Let be given. By the differentiability of at and the continuity of the reciprocal, there exists a such that implies

By the continuity of , there is a such that implies

Thus, for all , we have

as desired.

Finally, what if is complex differentiable? Complex differentiable functions often behave better than real differentiable functions, and the inverse function is no different.

Proposition 3 Let , where open, be a one-to-one holomorphic function. Let and set . Then , is holomorphic at and, moreover,

Proof: We prove by contradiction. Assume . The zeros of a holomorphic function are isolated, so we can pick an such that and are nonzero on the punctured disk . Let denote the order of the zero of at . since . Let denote the cicular path of radius about . By the argument principle,

Let denote the composition of of with . Then we have

where the right hand side denotes the widing number of about . Since is disjoint from and is closed, there is a such that the neighborhood is disjoint from . The winding number is constant on this neighborhood. Therefore, for any , we have

The left hand side is the number of zeros (with multiplicity) of . However, each zero of has multiplicity 1 since if then . Thus is not injective, a contradiction. Under the assumption , the proof in the real case goes through in in the complex as written.