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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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    Neither is the fundamental groupoid.2012-11-07

1 Answers 1

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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). $

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    So, this assumes we're computing the $\Bbb{Z}$-homology, yes?2012-11-06