Proving homeomorphism on closed ball Let $B$ be the closed unit ball in $\mathbb{R^n}$, and $a\in B,|a|<1$. Prove that $f:B\rightarrow B$,
$$f(x) = (1-|x|)a+x$$
Is an homeomorphism.
To me it's clear $f$ is continuous and I managed to prove it's injective. I'm only having trouble proving it's surjective, or finding an inverse. Any tips are appreciated!

I came up with an answer based on Proving a particular function is surjective on a Banach space. Notice that $g = f-Id$ is such that:
$$|g(x)-g(y)| = |(1-|x|)a-(1-|y|)a|=|a|||y|-|x||<||y|-|x||\leq|x-y|$$
So $g$ is a contraction. Now, for $y \in B$, put $h_y(x) = y-g(x)$. Then 
$$|h_y(x)-h_y(z)| = |g(z)-g(x)|<|x-z|$$
So again $h_y$ is a contraction, for any $y\in B$. Since B is closed and thus complete, by Banach's fixed point Theorem,  $\exists! x \in B$ such that $h_y(x)=x$, that is:
$$h_y(x)=x\iff y-g(x) = x\iff y -f(x)+x = x\iff f(x)=y$$
Thus $f$ is surjective. We conclude $f$ is bijective, so  As Math_QED pointed out, it's an homeomorphism.
