Spivak's Calculus on Manifolds - Proof of Inverse Function Theorem

Question

I have a small confusion in a step in the proof of the Inverse Function Theorem from Spivak's Calculus on Manifolds.

Theorem 2-11 (Inverse Function Theorem) Suppose that $f : \mathbb{R}^n \to \mathbb{R}^n$ is continuously differentiable in an open set containing $a$, and $\det f'(a) \neq 0$. Then there is an open set $V$ containing $a$ and an open set $W$ containing $f(a)$ such that $f : V \to W$ has a continuous inverse $f^{-1} : W \to V$ which is differentiable and for all $y \in W$ satisfies $$ (f^{-1})'(y) = (f'(f^{-1}(y)))^{-1}. $$

To prove this, we first reduce it to the case when $Df(a)$ is the identity map. Then, we find a closed rectangle $U$ containing $a$ in its interior which satisfies the following properties:

1. $f(x) \neq f(a)$ for $x \in U$.
2. $\det f'(x) \neq 0$ for $x \in U$.
3. $|D_j f^i(x)-D_j f^i(a)| < 1/2n^2$ for all $i$, $j$, and $x \in U$.
4. $|x_1 - x_2| < 2|f(x_1) - f(x_2)|$ for $x_1, x_2 \in U$.

There is also an open ball $W$ around $f(a)$ such that for all $y \in W$ and $x \in \operatorname{boundary} U$, we have

5. $|y - f(a)| < |y - f(x)|$.

Next, we will show that for any $y \in W$ there is a unique $x$ in $\operatorname{interior} U$ such that $f(x) = y$. To prove this, consider the function $g : U \to \mathbb{R}$ defined by $$ g(x) = |y-f(x)|^2 = \sum_{i=1}^n (y^i - f^i(x))^2. $$ This function is continuous and therefore has a minimum on $U$. Since $g(a) < g(x)$ for $x \in \operatorname{boundary} U$ by (5), the minimum occurs in the interior. So, there is a point $x$ in the interior of $U$ such that $D_j g(x) = 0$ for all $j$, that is $$ \sum_{i=1}^n 2(y^i - f^i(x))\cdot D_j f^i (x) = 0 \quad \text{for all $j$}. $$ By (2.) the matrix $(D_j f^i (x))$ must have non-zero determinant. Therefore we must have $y^i - f^i(x) = 0$ for all $i$, that is $f(x) = y$. This proves the existence of $x$. Uniqueness follows immediately from (4).

Then, Spivak says

If $V = (\operatorname{interior} U) \cap f^{-1}(W)$, we have shown that the function $f : V \to W$ has an inverse $f^{-1} : W \to V$.

My confusion is that since we have already shown that for all $y \in W$ there is a unique $x$ in the interior of $U$, why is it necessary to take $V = (\operatorname{interior} U) \cap f^{-1}(W)$? Doesn't $V = f^{-1}(W)$ suffice?

Answer 1

We have shown that for every $y \in W$ there is a unique $x \in \operatorname{interior} U$ such that $f(x) = y$. But, it could very well be the case that there is also some $x' \in \mathbb{R}^n \setminus U$ such that $f(x') = y$. To account for this case, we take $V = (\operatorname{interior} U) \cap f^{-1}(W)$.