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?

  • 2
    $\begingroup$ The homology groups of a space $X$ certainly are invariants of the isomorphism class $X$; in fact, they are homotopy type invariants. What specifically leads you to believe that homology is not an invariant under isomorphism? $\endgroup$ Commented Mar 24, 2019 at 14:00
  • $\begingroup$ I mean, you can't change the ker and image to any groups that they are isomorphic to and then form the homology group by taking the quotient. My professor confirmed this I didn't understand why. $\endgroup$
    – user637978
    Commented Mar 24, 2019 at 14:02
  • $\begingroup$ The homology groups are invariants of the isomorphism type of the chain complex given to you by, say, singular homology. The point is you need the isomorphisms to commute with the differentials. $\endgroup$
    – Tyrone
    Commented Mar 24, 2019 at 14:05

2 Answers 2


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.


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.


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