Question:
Suppose that $m < n$, that $U$ is an open set in $\mathbb R^m$ and that $f : U \rightarrow \mathbb R^n$ is a $C^1$ function that has rank $m$ everywhere in $U$. Show that for every $x$ in $U$ there is a neighborhood $V$ of $f(x)$ and a $C^1$ function $\tilde f : V \rightarrow \mathbb R^m$ that is a left inverse of $f$ in $V$ (that is, $\tilde f\circ f$ is the identity on $V$).
I do not understand the last sentence of this question. What does it mean by identity on $V$? Shouldn't the domain of $\tilde f\circ f$ is at least in $\mathbb R^m$ instead of $V$ which is in $\mathbb R^n$? Which of the following two is correct, please?
- $(\tilde f\circ f)(x^1,\cdots, x^m)=(x^1,\cdots, x^m)$; or
- $(\tilde f\circ f)(x^1,\cdots, x^m)=(x^1,\cdots, x^m, 0, \cdots, 0)$.
In addition, I started with trying constant rank theorem, but it did not seem to work since rank theorem only change coordinates in $\mathbb R^m$ and $\mathbb R^n$ without connecting them. Any hint, please?
Hint: consider the composition of $f$ with the projection onto the tangent space at $f(x)$ (considered as an $m$-dimensional affine subspace of ${\bf R}^n$).