proof that an arbitrary homeomorphism $h: B_{1}[0] \rightarrow B_{1}[0]$ maps $S^n$ to $S^n$ Intuitively this proposition seems true, but I've been told that is not a trivial thing to prove.
Is there any simple proof (or counter-example) for the proposition:

Consider the closed ball of radius 1: $B_{1}[0] \subset \mathbb{R}^{n+1}$ 
Given a homeomorfism $h: B_{1}[0] \rightarrow B_{1}[0]$, we have that $h(S^n) = S^n$

Is it easy to prove or disprove using basic topological arguments (if so, how would the proof go) or do we need to use tools such as Homology Theory?
 A: Start with the fact that for any homeomorphism of topological spaces $f:X \to Y$ and for any $x \in X$ and $y=f(x) \in Y$, the restricted function $f : X-\{x\} \to Y-\{y\}$ is a homeomorphism using the subspace topologies.
So for your problem it suffices to show that if $X=Y=B_1[0]$ and $x \in S^n$ and $y \in B_1[0]-S^n$ then $B_1[0] - \{x\}$ is not homeomorphic to $B_1[0]-\{y\}$. Furthermore, it suffices to prove that they are not homotopy equivalent.
First, $B_1[0]-\{x\}$ is contractible. In fact, it is star convex with respect to any of its points.
Second, $B_1[0]-\{y\}$ is homotopy equivalent to $S^n$, in fact it deformation retracts to $S^n$. A deformation retraction $f : B_1[0]-\{y\} \to S^n$ is defined for any $p \in B_1[0]-\{y\}$ by taking $f(p)$ to be the intersection of the ray $\{y + t(p-y) \,\bigm|\, t \ge 0\}$ with $S^n$.
So it suffices to prove that $S^n$ is not contractible. You can do this with homology, by computing $H_n(S^n;\mathbb{Z})=\mathbb{Z}$, whereas since $n \ge 1$, $H_n$ of any contractible space is trivial. (For the $n=0$ case, a connectivity argument does the job, or the homology argument can be adapted using reduced homology but that amounts to pretty much the same connectivity argument dressed up in a fancy costume).
I don't think there's any way around this basic setup to prove your statement. The argument can be localized, which leads to adaptations and generalizations. For instance a homeomorphism between manifolds must preserve the boundaries. Also, the invariance of domain theorem is a powerful theorem using the same ideas. Roughly speaking, somewhere in the proof, perhaps buried in an application of invariance of domain, there will be a calculation of $H_n(S^n;\mathbb{Z})=\mathbb{Z}$.
A: One possibility is to use the Invariance of Domain theorem (which in turn relies on non-trivial topology). 
Let $h\colon B_1\to B_1$ be a homeomorphism. Here $B_1$ is the closed unit ball in $\mathbb{R}^n$. Let $\mathring{B}_1$ denote the open unit ball.
The restriction of $h$ to $\mathring{B}_1$ is an injective continuous map of a domain of $\mathbb{R}^n$ into $\mathbb{R}^n$; therefore, the Invariance of Domain theorem applies and tells us that, for any $x\in\mathring{B}_1$, one has that $h(x)\in\mathring{B}_1$. This applies to $h^{-1}$ as well, and so $\left.h\right|_{\mathring{B}_1}$ is a homeomorphism of $\mathring{B}_1$ onto itself. 
The dual statement is that the restriction of $h$ to $S=B_1\setminus \mathring{B}_1$ is also a homeomorphism of $S$ onto itself, which is what we wanted to prove.
