# Rank theorem implies inverse function theorem

I am studying analysis on $\mathbb{R}^n$ and there is this question I cannot solve. Indeed it was not asked to me in any sense, but is usual to hear people saying that rank theorem is also one of the equivalents theorem to the inverse function theorem.

I have the following version of the rank theorem, and I am searching a way to prove that indeed, it implies the inverse function theorem.

Let $f : U \subset \mathbb{R}^{n+m} \to \mathbb{R}^{n+k}$ a $C^r(U)$ function with constant rank $n$, $U$ open set.

Then for each $z = (x,y) \in U$ there are diffeomorphisms $\alpha$ defined on one open neighborhood of $f(z)$ to an open set containing $(x,0)$ and $\beta$ defined on one open set of $\mathbb{R}^{n+m}$ to another containing $z$ such that $\alpha \circ f\circ \beta(x,y) = (x,0).$

I know that if I can argue that if $f : \mathbb{R}^n \to \mathbb{R}^n$ and $f$ has constant rank $n$ then $\alpha = id$ the identity, the function $\beta$ will work as a candidate for the inverse and then I can finish showing that the rank theorem implies the inverse function theorem, but I don't know how to proceed.

Thanks a lot!

• It seems like you answered your question. Are you just confused about how to go from full-rank-at-a-point to full-rank-in-an-open-neighborhood? – Tim kinsella May 16 '16 at 6:12
• @Timkinsella, I am confused indeed with the fact, can I have a good reason to state that $\alpha$ is the identity? – L.F. Cavenaghi May 16 '16 at 14:42
• So you know $\alpha f \beta$ is a diffeo and $\alpha, \beta$ are diffeos. So what can you conclude about $f$? – Tim kinsella May 16 '16 at 22:46
• thank you so much!!!!!!!!!!!!!!!!!!!!!!! @Timkinsella – L.F. Cavenaghi May 16 '16 at 23:18