I really need a help with the following exercise:
Suppose $\mathbb{E}$ and $\mathbb{F}$ are normed spaces, $A \subseteq \mathbb{E}$ is an open set, $f: A \to \mathbb{F}$ is differentiable on $A$, and suppose that $g: A \to A \times \mathbb{F}$ and $s: A \times \mathbb{F} \to \mathbb{F}$ are functions defined by $g(x)=(x,f(x))$ and $s(x,z)=z-f(x)$.
- Prove that $g$ and $s$ are differentiable on $A$ and $A \times \mathbb{F}$, respectively.
- Prove that $g'(x): \mathbb{E} \to \mathbb{E} \times \mathbb{F}$ is injective.
- Prove that the image of $g'(x)$ is the kernel of $s'(x,z)$.
I have proven the item 1, but I'm not sure about item 2. However I think it can be proved using the fact that a linear transformation $T$ is injective if and only if $Ker(T)=\{0\}$. I have no idea how to prove item 3. Any hint will be greatly appreciated.
(2.) If $g$ is injective then $g'$ is injective. Notice $g(x_1)=g(x_2)$ implies $(x_1,f(x_1)) = (x_2,f(x_2))$ hence $x_1=x_2$ which shows $g$ is injective.
(my notation below may be non-standard, but, I believe when you clean-up the solution it's something like what follows:)
(3.) Suppose $s(x,z)=z-f(x)$. Observe $g(x) = (x,f(x))$ implies $g'(x) = Id \times f'(x)$. Then $s'(x,z) = -f'(x) \times Id$. But, $s'(g'(x)) = f'(x)-f'(x)=0$ thus $g'(x)$ is in the kernel of $s'$.