Homeomorphic spaces have the same homology groups How do I show that homeomorphic spaces have isomorphic homology groups?
Hatcher says that it is evident from the definitions, which makes me think that I didn't understand something. How does a homeomorphism between two spaces $X$ and $Y$ induce an isomorphism between $\ker{∂_n}/\operatorname{Im}∂_{n+1}$ and $\ker{∂'_n}/\operatorname{Im}∂'_{n+1}$ ($∂$'s are the boundary maps here)? I would send the equivalence class $\operatorname{Im}∂_{n+1}$ to the class $\operatorname{Im}∂'_{n+1}$ but where do I send the classes of the other elements of $\ker{∂_n}$?
 A: Hint: Let $f:X\rightarrow Y$ be a homeomorphism, and $g:Y\rightarrow X$ its inverse. Let $f_*,g_*$ be the maps induced in homology. Then you can easily show that $f_*$ and $g_*$ are inverse one of the other.
A: Here is a categorical explanation. (Doing algebraic topology without using categories is not the best.)
One usually defines singular homology as the composite functor 
$$ \mathbf{Top} \to [\Delta^{\mathrm{op}},\mathbf{Set}] \to [\Delta^{\mathrm{op}},R\!-\!\mathbf{Mod}] \to \mathfrak{Ch}(R\!-\!\mathbf{Mod}) \to R\!-\!\mathbf{Mod},$$
where


*

*the first arrow is the simplicial set of singular simplexes functor : $X \mapsto \hom_{\mathbf{Top}}(\Delta_\bullet, X),$

*the second arrow is the free $R$ module functor on each set of the simplicial set : $S_\bullet \mapsto R[S_\bullet]$,

*the third one is the singular chain complex functor : $M_\bullet \mapsto (M_n, d_n = \sum_i (-1)^i \partial_i^n)_{n \in \mathbb N}$,

*and the last one is the functor homology of chain complexes : $(M_n,d_n) \mapsto \ker d_n\,\big/\,\mathrm{im}\, d_{n+1}$.


So the simple fact that this is a functor (as a composition of functors) gives that it preserves isomorphisms : hence an isomorphism of $\mathbf{Top}$ (i.e. a homeomorphism) is mapped to an isomorphism of $R\!-\!\mathbf{Mod}$ (i.e. an isomorphism of $R$-modules).
A: Using the notation found in Vick's Homology Theory.
Let $f: X \to Y$ be a homeomorphism.  Let $\sigma_n = \{ (s_0, s_1, \dots, s_n) \in \Bbb{R}^{n+1} : \sum_{i=0}^n s_i = 0$ and $s_i \geq 0, \ \forall i \}$.  That is the standard $n$-simplex.
Then if $\phi : \sigma_n \to X$ is singular $n$-simplex, then $f\circ \phi : \sigma_n \to Y$ is continuous since $f$ is by assumption and $\phi$ is by definition of singular $n$-simplex, and so $f \circ \phi$ is a singular $n$-simplex into $Y$.
Let $S_n(X), S_n(Y)$ be the free abelian groups on singular $n$-simplices into $X$ and into $Y$ respectively.
Then $f_* : S_n(X) \to S_n(Y)$ given by $f_*(\phi) = f\circ \phi$ extended (uniquely) homomorphically to all of $S_n(X)$ has inverse $f_*^{-1} : S_n(Y) \to S_n(X)$ given by $f_*^{-1}(\psi) = f^{-1}\circ \psi$.
Proof of inverse:  $x = \sum_{i=1}^r x_r \phi_r \in S_n(X)$ where $x_r \in \Bbb{R}, \ \forall r$, a general element, then $f_*(x) = \sum_{i=1}^r x_r f \circ \phi_r$.  Clearly applying $f_*^{-1}$ to that expression will undo it back to $x$.
$f_*, f_*^{-1}$ are both abelian group homomorphisms because they were constructed via extending the maps on the basis of singular $n$-simplices $\phi$, homomorphically, and that process always results in a homomorphism.
Therefore, we've proven that $S_n(X) \simeq S_n(Y)$ whenever $f : X \to Y$ is a homeomorphism.
$$
\require{AMScd}
\begin{CD}
S_{n+1}(X) @>\partial>> S_{n}(X) @>\partial>> S_{n-1}(X)\\
@Vf_*VV @Vf_*VV @Vf_*VV\\
S_{n+1}(Y) @>\partial'>> S_{n}(Y) @>\partial'>> S_{n-1}(Y)
\end{CD}
$$
Let $\partial_i, \partial_i'$ be the $i$th face maps of $\partial, \partial'$ where $\partial := \sum_{i=0}^n (-1)^i \partial_i$ is the definition of the boundary map.
Let $\phi$ be a singular $n$-simplex in $S_n(X)$, i.e. a basis element for the free abelian group that is $S_n(X)$.
Clearly, it is sufficient to show that $f_*\partial_i(\phi) = \partial_i'f_*(\phi)$ for any $i = 0..n$, and then everything should extend homomorphically and so then all squares in the above boundary map diagram will then be proven commutative.
$$
f_* \partial_i(\phi)(s_0, \dots, s_{n-1}) = f \circ \phi(s_0, s_1, \dots, s_{i-1}, 0, s_i, \dots, s_{n-1})
$$
Clearly doing $f\circ \phi$ first and then applying the $\partial_i$ operator will result in the same thing!  This should be easy for the reader to see by now.  Or, simply note that $f\circ \phi = \phi'$ is just any other singular $n$-simplex so the above formula comes out the same when you apply $\partial_i$ after applying $f_*$.
We clearly then have an isomorphism of chain complexes $S_*(X), S_*(Y)$!

Next, we are going to prove that two isomorphic chain complexes have isomorphic $n$th homology groups.
Let $Z_n = Z_n(X)$ be the group of $n$-cycles and $B_n = B_n(X)$ the group of $n$-boundaries for the topological space $X$.
Then $x \in Z_n \implies \partial(x) = 0$ so that $f_*\partial(x) = 0$, but $f_*\partial(x) = \partial f_*(x)$ so that $f_*(x) \in \ker \partial$ (on the $Y$ side now).  And so $f_*$ restricted to $Z_n$ takes $Z_n(X)$ to $Z_n(Y)$.
I will pause and let the reader prove a similar situation for $f_*$ taking $B_n(X)$ to $B_n(Y)$.
Thus define a map from $H_n(X) = Z_n(X) / B_n(X)$ into $H_n(Y)$ by:
$$
h: x + B_n(X) \mapsto f_*(x) + B_n(Y)
$$
You need to show two things: that $h$ is an abelian group homomorphism and that it is well-defined.  Any time you have a map involving a quotient group you should automatically always check for well-definedness, or in other words that $x + B_n = y + B_n \implies h(x) = h(y)$.  Two more exercises for the reader.
Now, clearly $f_*^{-1}$ being the inverse homomorphism going the other direction, defining and applying an analogous $h^{-1}$ will undo $h$.  But you already proved that $h$ is well-defined and a homomorphism, so that you do not have to do the same for $h^{-1}$, simply arguing by problem symmetry.
And so we can now safely conclude that $H_n(X) \simeq H_n(Y)$ whenever $f: X \to Y$ is a given homeomorphism of topological spaces.  $\square$
