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!
 A: Hint: Continuous bijection from a compact space to a Hausdorff space is a homeomorphism.
A: This answer uses some light algebraic topology (really, just the definitions of homotopy, contractible) to show that $f$ is surjective. From there, showing $f$ is a homeomorphisn is straightforward.
Note first that $f$ preserves the unit circle (or, if you prefer, the boundary of the closed ball, i.e. its difference with the open ball). Note also that the unit ball is contractible. Assume for contradiction that $f$ misses a point $x$. Since $f$ preserves the unit circle, $im(f)$ is not contractible (think about where $x$ is moved to under a homotopy). But $im(f)$ is homeomorphic to a contractible space, the unit ball, and therefore must itself be contractible, a contradiction.
A: 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.
A: Suppose $|y|=1.$ Then for $t\in [0,1],$ $f(ty) = (1-t)a+ ty.$ Thus $f$ maps the line segment $[0,y]$ onto the line segment $[a,y].$ Therefore
$$f(B) = \cup_{|y|=1}f([0,y]) = \cup_{|y|=1}[a,y].$$
It's straightforward to verify that the set on the right is all of $B.$ Therefore $f$ is surjective.
