Proving a particular function is surjective on a Banach space. Let $(E,\|x\|)$ and let $f: E \to E$ such that $f+Id$ is a contraction ($Id$ is the identity map). Prove that $f$ is surjective and prove that, if $f$ is linear, then $f$ is a homeomorphism. 
The attempt at a solution:
I am having a hard time with this exercise. For the first part, I want to prove that given $y \in E$, there exists $x \in E$: $f(x)=y$. If I define the function $g(x)=f(x)+Id$, then $g$ must have a unique fixed point by the fixed point theorem. However, I don't know how to use this information to find the $x$ such that $f(x)=y$ for a given $y$.
For the second part, I am confused about the following: suppose I've proved $f$ is surjective, would this imply $f$ is injective? I know this is true for endomorphisms defined on a finite dimensional vector space, but $E$ could be of infinite dimension. So, how could I assure that $f^{-1}$ exists?. I also have problems to prove that $f$ is continuous.
I would appreciate if someone can help me (with hints, ideas or suggestions) to prove these things.
 A: After the suggestions, I could solve some parts of the problem:
To prove $f$ is surjective, given $y \in E$, if one defines $h_y:E \to E : h(x)=g(x)-y$, then for $x,z \in E$, $d(h(x),h(z))=d(g(x)-y,g(z)-y)=d(g(x),g(z))<d(x,z)$, which means $h_y$ is a contraction on a complete metric space, then $h_y$ has a unique fixed point. Let $x$ be the fixed point of $h_y$, $h(x)=x \iff g(x)-y=x \iff f(x)+x-y=x \iff f(x)=y$, it follows that $f$ is surjective. 
For the second part, suppose $f$ is linear. We have to prove that $f$ is continuous, bijective and that $f^{-1}$ is also continuous. As $E$ is a Banach space, $f$ is continuous $\iff$ $f$ is bounded. 
$\|f(x)\|=\|f(x)+Id(x)-Id(x)\|\leq \|f(x)+Id(x)\|+\|Id(x)\|$
By hypothesis, $f+Id$ is a contraction $\implies$ $f+Id$ is continuous $\implies$ $f+Id$ is bounded. Then, for every $x \in E$, $\|f(x)\|\leq \|f(x)+Id(x)\|+\|Id(x)\|\leq c\|x\|+\|x\|=(c+1)\|x\|$, this proves $f$ is continuous. 
With Stephen's comments I could prove injectivity of $f$ and continuity  of $f^{-1}$:
First lets prove that there exists $m>0 : m\|x\|\leq \|f(x)\|$:
$\|x\|=\|x+f(x)-f(x)\|\leq \|x+f(x)\|+\|f(x)\| \implies \|x\|-\|x+f(x)\|\leq \|f(x)\|$. We know there exists $c>0:\|x+f(x)\|\leq c\|x\| \implies (1-c)\|x\|\leq \|x\|-\|x+f(x)\|\leq \|f(x)\|$. 
Suppose there are $a\neq b, a,b \in E: f(a)=f(b) \iff f(a)-f(b)=0 \iff f(a-b)=0$.
But $0<m\|a-b\|\leq \|f(a-b)\|=0$, which is absurd.
So there exist $m,M>0: m\|x\|\leq \|f(x)\|\leq M\|x\|$, with this, I can prove that $f^{-1}$ is bounded:
Let $y \in E$, then there is $x \in E: y=f(x) \implies m\|f^{-1}(y)\|=mf^{-1}(f(x))=m\|x\|\leq \|f(x)\|=\|y\|$. Then $\|f^{-1}(y)\|\leq \dfrac{1}{m}\|y\| \implies f^{-1}$ is continuous. It follows that $f$ is a homeomorphism.
