For every , there is an with . By Theorem 2-11, there is an open set and an open subset such that and . Since clearly , this shows that is open. Furthermore is differentiable. It follows that is differentiable at . Since was arbitrary, it follows that is differentiable.
By applying the previous results to the set in place of , we see that is open.
We will show the result is true even if is only defined in a non-empty open subset of . Following the hint, we know that is not constant in any open set. So, suppose we have (the case where is analogous). Then there is an open neighborhood of with for all . The function defined by satisfies for all . Assuming that and hence are 1-1, we can apply Problem 2-36. The inverse function is clearly of the form and so for all . Now is open but each horizontal line intersects at most once since is 1-1. This is a contradiction since is non-empty and open.
By replacing with a vector of variables, the proof of part (a) generalizes to the case where is a function defined on an open subset of where .
For the general case of a map where is an open subset of with , if is constant in a non-empty open set , then we replace with and drop out reducing the value of by one. On the other hand, if for some , then consider the function defined by . Just as in part (a), this will be invertible on an open subset of and its inverse will look like . Replace with . Note that we have made . Again, by restricting to an appropriate rectangle, we can simply fix the value of and get a 1-1 function defined on on a rectangle in one less dimension and mapping into a space of dimension one less. By repeating this process, one eventually gets to the case where is equal to 1, which we have already taken care of.
Suppose one has for some . By the mean value theorem, there is a between and such that . Since both factors on the right are non-zero, this is impossible.
Clearly, for all . The function is not 1-1 since for all .
to show that continuity of the derivative cannot be eliminated from the hypothesis of Theorem 2-11.
Clearly, is differentiable for . At , one has
So satisfies the conditions of Theorem 2-11 at except that it is not continuously differentiable at 0 since for .
Now and it is straightforward to verify that for all sufficiently large positive integers . By the intermediate value theorem, there is a between and where . By taking n larger and larger, we see that is not 1-1 on any neighborhood of 0.