Inverse Function Theorem as stated in Milnor's Topology from the Differentiable Viewpoint

Question

Inverse Function Theorem If the derivative $df_x : \mathbb{R}^k \to \mathbb{R}^k$ is nonsingular, then $f$ maps any suffiiciently small open set $U'$ about $x$ diffeomorphically onto an open set $f[U']$

Now my question is what is rigorously meant by suffciently small open set? Certainly $U'$ can't be any open set, since if $U' = \mathbb{R}^k$, that would need the extra assertion that $f$ must be bijective.

Since $U'$ is open in $\mathbb{R}^k$, $U'$ must be the open ball centered at $x$ with radius $\epsilon$, thus $U' = B_{(\mathbb{R}^k, d)}(x, \epsilon)$ for some $\epsilon > 0$. (and $d$ is that standard euclidean metric).

"Sufficiently small" in the above theorem must mean that there is some upper bound on $\epsilon$, if that's correct, then what would that upper bound be?

Answer 1

If $U'$ is any such open ball around $x$ of radius $\epsilon$, then $f$ maps maps any open ball around $x$ of radius $\delta<\epsilon$ diffeomorphically onto its image as well. So "sufficiently small" in this case means "smaller than $\epsilon$", once you have some value of $\epsilon$ that works.

Answer 2

You misquoted Milnor. He wrote “open set $U'$ about $x$” and not “open set $U'$ of $x$”.

It is false that $U'$ must be an open ball centered at $x$. It doesn't even have to be an open ball at all.

The meaning is this: there is a $\varepsilon>0$ such that, for any open set $U'$ such that $x\in U'$, $f$ maps $U'\cap B(x,\varepsilon)$ diffeomorphically onto $f\bigl(U'\cap B(x,\varepsilon)\bigr)$.