# What happens to homology without cycles?

Let X be a topological space and $C_n(X)$ be the singular chain complex. The homology is defined to be $H_n(X)$ = $ker \partial_n / im \partial_{n+1}$.

What happens if we take $K_n(X) = C_n(X) / im \partial_{n+1}$ instead?

(The idea comes from comparing the fundamental group to the fundamental groupoid)

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Have you tried to compute such thing, for example? Try and see for yourself what happens. –  Grigory M Nov 6 '12 at 22:04
I don't understand the downvote. The question is well-formulated and it's an interesting idea which is motivated by analogy with another, better-known interesting idea. It's certainly not a homework problem. –  user29743 Nov 6 '12 at 22:07
It's not invariant under homotopy. –  Qiaochu Yuan Nov 6 '12 at 23:03
Neither is the fundamental groupoid. –  user29743 Nov 7 '12 at 5:41

Write $L_n(X)$ for chains mod cycles. We have a short exact sequence
$$0 \to H_n(X) \to K_n(X) \to L_n(X) \to 0$$
On the other hand, $d$ induces an inclusion $$0 \to L_n(X) \to C_{n-1}(x),$$ i.e. $L_n$ is a submodule of a free, so is free, in fact is isomorphic to $B_{n-1}(X).$ Thus our sequence splits (though not naturally in $X$) and $$K_n(X) = B_{n-1}(X) \oplus H_n(X).$$
Thanks countinghaus. Do you think there's any chance of a 'geometric' interpretation of this? I'm thinking about $H_1(X)$ in particular –  krey Nov 6 '12 at 22:36
So, this assumes we're computing the $\Bbb{Z}$-homology, yes? –  Kevin Carlson Nov 6 '12 at 23:02