Let $C_q(K)$ be a group of $q$-chains on a given simplicial complex $K$. Since this is a free abelian group, its subgroup must be a free abelian group, especially $Z_q(K),B_q(K)$. Then we define the homology group $H_q(K)$ as $Z_q(K)/B_q(K)$. And here is a theorem:

Let $G,G_1,G_2$ be a modules. If $G\cong G_1\oplus G_2$, then $G/G_2\cong G_1$.

We can prove this easily. Corresponding $g+G_2$ to $g$, it indicates isomorphism.

As the passage above says, $Z_q(K)$ is some products of integers $Z$, and so is $B_q(K)$. And $r(B_q(K))\leq r(Z_q(K))$ since $B_q(K)$ is a subgroup of $Z_q(K)$, where $r(G)$ means the rank of abelian group $G$. For example, $Z_q(K)\cong Z\oplus Z$, $B_q(K)\cong Z$ will happen.

So $H_q(K)$ seems to be a free abelian group. For the example above, $H_q(K)\cong (Z\oplus Z)/Z \cong Z$ by the theorem. But I know this is not true. For instance, $H_1(K)\cong Z\oplus Z_2$ on the Klein Bottle. Could someone tell me where I was wrong? Thanks.


The issue is that a quotient of free abelian groups need not be free. Take, for instance, the quotient of $\mathbb{Z}$ by the subgroup $2\mathbb{Z}$. The quotient is $\mathbb{Z}_2$, and this has torsion.


Another point to what Peter Crooks said: Your statement that if $G \cong G_1 \oplus G_2$ then $G/G_2 \cong G_1$ is not necessarily correct! It depends on what isomorphisms and embeddings that we take.

If we consider $G_1 \cong G_2 \cong \mathbb{Z}$ and $G \cong \mathbb{Z} \oplus \mathbb{Z}$, then it is certainly true that $G \cong G_1 \oplus G_2$. However, if we embed $G_2 \hookrightarrow G$ as the subgroup

$$ G_2 = \{(n,2n) \in G\} $$

then the quotient is not isomorphic to $G_1$.

  • $\begingroup$ I got your explanation. My statement will be true at least if $G$ can be decomposed as $G\cong G_1\oplus G_2$ with direct sum. In other words, $G_1$ and $G_2$ embed in $G$ by inclusion. $\endgroup$ – Mike Mar 20 '14 at 17:14
  • $\begingroup$ So, don't we calculate $H_q(K)$ simply like my last example? That is, $Z_q(K)\cong Z\oplus Z$ and $B_q(K)\cong Z$ then $H_q(K)\cong Z$. $\endgroup$ – Mike Mar 20 '14 at 17:25
  • $\begingroup$ It is not enough to know that $Z_q(K) \cong \mathbb{Z} \oplus \mathbb{Z}$ and that $B_q(K) \cong \mathbb{Z}$ abstractly, you need to know how $B_q(K)$ is embedded into $Z_q(K)$. If it helps, remember that $2\mathbb{Z} \cong \mathbb{Z}$ as groups, but $\mathbb{Z} / 2\mathbb{Z} \ncong \mathbb{Z} / \mathbb{Z}$. $\endgroup$ – Simon Rose Mar 20 '14 at 18:31

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