Converting topological problems to algebraic problems We can convert topological problems to algebraic problems by functors. Can you give  any example about this situation?
 A: Consider the question:

Are $\mathbb{R}^n$ and $\mathbb{R}^m$ homeomorphic for $n\neq m$?

When $n=1$ and $m>1$, the answer is not too difficult, since $\mathbb{R}^m\setminus\{0\}$ is connected for $m>1$ while $\mathbb{R}\setminus\{0\}$ is not.  However, for general $n,m$, the question is much more difficult to handle.
One of the main ideas of Algebraic Topology is using "algebraic invariants".  Formally, we wish to build functors from some sufficiently nice subcategory of $\mathbf{Top}$, like the category of compactly-generated hausdorff spaces $\mathbf{CGHaus}$ (this is because $\mathbf{Top}$ has some poor properties as a category), into some "algebraic" category like $\mathbf{Grp}$ or $\mathbf{AbGrp}$.  
For example, there is a functor $\pi_1:\mathbf{Top}_\bullet \rightarrow \mathbf{Grp}$ given by sending a pointed topological space $(X,x_0)$ to the fundamental group $\pi_1(X,x_0)$.  A slightly simpler example is the functor $\pi_0:\mathbf{Top}\rightarrow \mathbf{Set}$ defined by sending a topological space $X$ to $\pi_0 X$, the set of path-connected components of $X$.
Why do we do this?  We do this because if, given two categories $\mathscr{C},\mathscr{D}$ and a functor $F:\mathscr{C}\rightarrow \mathscr{D}$, then if there exists an isomorphism $\varphi:c\rightarrow c'$ between the objects $c$ and $c'$ of $\mathscr{C}$, then $F(\varphi):F(c)\rightarrow F(c')$ is also an isomorphism.  As such, if $F(c)$ and $F(c')$ aren't isomorphic, then we can conclude that $c$ and $c'$ aren't isomorphic!
As far as our specific question above is concerned, the algebraic invariant we want to make use of is that of simplicial homology.  The category $\mathbf{Csim}$ is that of simplicial complexes, and $\mathbf{Ch}_\bullet(\mathbf{AbGrp})$ the category of chain complexes over $\mathbf{AbGrp}$.  Then we can send a simplicial complex $K$ to its simplicial homology chain complex, giving a functor.
Simplicial complexes aren't topological spaces, but there exists a functor $|\cdot|:\mathbf{Csim}\rightarrow \mathbf{Top}$ called the geometric realization, and it can be shown that if $|K|$ and $|K'|$ are homotopy equivalent, then the homology of $K$ is equal to the homology of $K'$.  Using this, we get a functor from the category of triangulable topological spaces (those that are homeomorphic to the realization of some simplicial complex) to $\mathbf{Ch}_\bullet(\mathbf{AbGrp})$.  
$\mathbb{S}^n$, the $n$-sphere, is triangulable and has homology given by
$$H_m(\mathbb{S}^n)=\begin{cases} \mathbb{Z} & i=0,n \\ 0 & \text{otherwise.} \end{cases}$$
So to finally apply all of this to our original question, suppose $\mathbb{R}^n\cong \mathbb{R}^m$.  Then we can show that $\mathbb{R}^n\setminus\{0\}\cong \mathbb{R}^m\setminus \{0\}$.  $\mathbb{R}^n\setminus \{0\}$ is homotopy-equivalent to $\mathbb{S}^n$, so our homeomorphism would descend to showing that $\mathbb{S}^n$ and $\mathbb{S}^m$ are homotopy-equivalent.  But the homologies of homotopy-equivalent spaces are equal, so if $n\neq m$ (and $m\neq 0$) we have $\mathbb{Z}=H_n(\mathbb{S}^n)=H_n(\mathbb{S}^m)=0$, giving a contradiction.  (If $m=0$, then just switch the role of $m$ and $n$.)  Thus, $\mathbb{R}^n$ cannot be homeomorphic to $\mathbb{R}^m$ when $n\neq m$.
A: Problem: Let $B^n\in\mathbb{R}^n$ denote the closed unit ball, and let $S^{n-1}$ denote its boundary. Show that there is no continuous map $f:B^n\to S^{n-1}$ such that$$f|_{S^{n-1}}=id.$$
