Why are homology groups not invariant under isomorphism classes? My classmates and I were calculating the first homology group of the klein bottle, and we saw that $Ker \, \delta_1 \cong \mathbb{Z} \oplus \mathbb{Z} \oplus \mathbb{Z}$ and $Im \, \delta_2 \cong \mathbb{Z} \oplus \mathbb{Z}$.
However, $H_1(K) \ncong \mathbb{Z}$. We got the correct answer, $H_1(K) \cong \mathbb{Z} \oplus \mathbb{Z_2}$ once we considered the group presentation where our $Im \, \delta_2$ were used as relations in the presentation.
But this led me to wonder why  homology groups are not invariant under isomorphism classes?
 A: Suppose that I have a chain complex $(C,\delta)$, i.e. a sequence of groups $\{C_n\}_{n\in \mathbb{Z}}$ and differentials $\delta_n: C_n \to C_{n-1}$ such that $\delta_{n-1}\delta_n = 0$ for all $n \in \mathbb{Z}$. Then indeed chain complexes with the same underlying groups may have different homology. For example, consider:
$$ \cdots \to 0 \to \mathbb{Z} \to_\varphi \mathbb{Z} \to 0 \to \cdots$$
If $\varphi$ is multiplication by $2$, then there is a homology group isomorphic to $\mathbb{Z}/2\mathbb{Z}$; if $\varphi$ is the identity, there is no nonzero homology. This phenomenon is not just limited to torsion; the ranks of homology groups of complexes may also differ.
The issue is that:

An isomorphism of underlying groups is not the right notion of isomorphism for chain complexes.

Homology is defined in terms of the differentials, as well as the underlying groups. A reasonable notion of isomorphism for a chain complex should somehow respect the differentials, so that homology is indeed an invariant of isomorphism of chain complexes. Here is the definition:
Definition: An isomorphism of chain complexes $(C,\delta)$ and $(C', \delta')$ is a sequence of maps $f_n: C_n \to C_{n}'$ which are isomorphisms and such that $\delta'_n f_n =f_{n-1} \delta_n$, that is, the following diagram commutes:
$$\require{AMScd}\begin{CD}
C_n @>f_n>>  C'_n \\
@V\delta_nVV @V\delta'_nVV \\
C_{n-1} @>f_{n-1}>> C'_{n-1}\\
\end{CD}$$
It is a nice exercise to show that if two chain complexes are isomorphic in the above sense, then they have isomorphic homology groups. This may also be generalized to the notion of a homomorphism of chain complexes; one of the beautiful elements of algebraic topology is that  (appropriate) maps on spaces induce maps on chain complexes. 
A: I think your question may be related to this more general question:
Isomorphic quotients by isomorphic normal subgroups
In general, even if $H\cong K$, it is not guaranteed that the quotient groups  $G/H$ and $G/K$ are isomorphic.
Since homology is defined to be the quotient group $\ker\partial/\text{Im}\partial$, by the above reasoning we can't just use "isomorphism classes" to conclude the final homology.
