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In Singular homology, let $C_n(X)$ be the free abelian group generated by all the $n$-siimplices of the topological space $X$. Let $U$ be a subspace of $X$, then we have a spliting sequence $0\rightarrow C_n(U) \rightarrow C_n(X) \rightarrow C_n(X,U) \rightarrow 0$.

But why we do not have the splitting $H_n(X) \cong H_n(U) \oplus H_n(X,U)$ of homologies?

Aubrey
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  • We just don't. What we have instead is the long exact sequence. – Qiaochu Yuan May 19 '15 at 04:41
  • A relative cycle in $C_n(X,U)$ may not be the projection of a cycle in $C_n(X)$. – Joe S May 19 '15 at 14:44
  • @JoeS I don't understand what you mean. It is true that the group $C_n(X,U)$ is the quotient of $C_n(X)$ by its subgroup $C_n(U)$, essentially by definition. – Najib Idrissi May 19 '15 at 16:10
  • @Najib Idrissi A cycle in $C_n(X,U)$ is the image of some chain in $C_n(X)$ but that chain may not be a cycle. What you do know is that the boundary of that chain projects to zero so it is the image of the inclusion of $C_{n-1}(U)$ into $C_{n-1}(X)$. This is how you get the connecting homomorphism and the ong exact homology sequence. – Joe S May 20 '15 at 19:59
  • Ah I see what you mean now @JoeS. I missed the word "cycle", I read "chain"... – Najib Idrissi May 21 '15 at 07:24

2 Answers2

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You're right that for any pair $(X, A)$, there is a split short exact sequence of chain complexes

$$0 \to C_\bullet(A) \to C_\bullet(X) \to C_\bullet(X, A) \to 0$$

But it is not true that having short exact sequences at the chain level implies that the snake maps $\partial$ at the homology levels are zero, i.e., you get a short exact sequence at the homology level. This is because a chain-level injective/surjective map might not induce an injective/surjective map at the homology level, see this question.

The point of the long exact sequence of homology is that it measures failure of short exactness of the $H_\bullet$ functors, to emphasize.

One a different note, if there is a retract $r : X \to A$, then the induced $H_\bullet(r) : H_\bullet(X) \to H_\bullet(A)$ acts as a left-inverse for the maps $H_\bullet(X) \to H_\bullet(A)$, making the snake maps vanish and becomes a section of the resulting short exact sequence

$$0 \to H_\bullet(A) \to H_\bullet(X) \to H_\bullet(X, A) \to 0$$

Which implies $H_\bullet(X) \cong H_\bullet(A) \oplus H_\bullet(X, A)$.

Community
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Balarka Sen
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Adding an answer here that I found helpful when thinking about why the homology groups split in the case of retractions: a retraction $r:X \to A$ is left inverse for the inclusion $i$. Functors always preserve mutual left/right inverse-ness (in other words they preserve split epi/mono pairs) so there is a left inverse for the induced map $H_\bullet(i):H_\bullet(A) \to H_\bullet(X)$. On the other hand, the pointwise splitting of a chain complex is not necessarily a chain-map. More precisely, the short exact sequence of chain groups $$ 0 \to C_n(A) \to C_n(X) \to C_n(X, A) \to 0$$ splits in $\mathrm{Ab}$ for any given $n$, but the short exact sequence of chain-complexes $$ 0 \to C_\bullet(A) \to C_\bullet(X) \to C_\bullet(X, A) \to 0$$ may not split in the category of chain complexes. If the latter sequence splits (such as when there is a retraction $r:X \to A$, or when $X$ is the direct sum of $A$ and some other space), then by applying the homology functor you do in fact get a splitting of the homology groups.

abhi01nat
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