# Inverse Function Theorem proof: f is injective

I am trying to prove the Inverse Function Theorem from the Implicit Function Theorem for Banach spaces. My attempt so far is as follows:

Let $$f:\mathbf{X}\to \mathbf{Y}$$ be a $$\mathcal{C}^k$$ function between Banach spaces, and let $$x^*\in\mathbf{X}$$ and $$y^*:=f(x^*)$$ be such that the Fréchet derivative $$\mathrm{d}\,f(x^*):\mathbf{X}\to\mathbf{Y}$$ is bounded with bounded inverse.

Consider $$F:\mathbf{X}\times\mathbf{Y}\to\mathbf{Y}$$ given by $$F(x,y):=f(x)-y$$. Then the partial Fréchet derivative $$\partial_xF(x^*,y^*)\equiv \mathrm{d}\,f(x^*)$$ is bounded with a bounded inverse, so $$F$$ satisfies the hypotheses of the Implicit Function Theorem. Hence, there are open sets $$U\subseteq\mathbf{X}$$ and $$V\subseteq\mathbf{Y}$$ containing $$x^*$$ and $$y^*$$, respectively, and a $$\mathcal{C}^k$$ function $$g:V \to U$$ such that $$F(g(y), y) = 0$$ for all $$y\in V$$, i.e. $$f(g(y)) = y$$.

The above shows that $$g$$ is a right inverse of $$f$$, but I want to be able to conclude, further, that $$g$$ is a left inverse of $$f$$ on $$U$$, i.e. $$g(f(x)) = x$$ for all $$x \in U$$. I know this to be equivalent to showing that the restriction $$f|_U$$ is injective (possibly by choosing a smaller neighbourhood of $$x^*$$ in $$U$$), but I am not sure how to do this.

I would greatly appreciate any hints!

• Doesn't the classical formulation of the implicit function theorem say $f:X\oplus Y \to Z$ etc – JJR May 16 '17 at 11:11
• @JJR Are you referring to my writing '$X \times Y$' instead of '$X \oplus Y$? The statement of the theorem as I know it has $F$ defined on an open set $W \subseteq X \times Y$ (I have taken $W = X \times Y$ for simplicity). – ryan221b May 16 '17 at 13:10
• Oh sry my mistake I thought you were proving implicit function theorem from inverse function theorem not the other way around – JJR May 16 '17 at 13:15