If $X$ is a space with a pair of subspaces $A, B \subset X$ such that $X$ is the union of the interiors of $A$ and $B$, then there is a long exact sequence of homology groups

$\displaystyle \ldots\to H_n(A\cap B) \quad \xrightarrow{\Phi}\quad H_n(A) \oplus H_n(B) \quad \overset{\Psi}{\to} \quad H_n(X) \quad \overset{\partial}{\to} \quad H_{n-1} (A \cap B)\to \ldots$

The explicit action of the boundary map $\partial : H_n(X)\to H_{n-1}(A \cap B)$ escapes me. Allen Hatcher, in his book, writes

A class $\alpha \in H_n(X)$ is represented by a cycle $z$, and by barycentric subdivision or some other method we can choose $z$ to be a sum $x+y$ of chains in $A$ and $B$, respectively.

I think I can understand how $\partial \alpha$ can then be identified with $H_{n-1}(A \cap B)$, but can someone make clear to me how we choose such a sum? If you can go through all the steps taking an element in $H_n(X)$ to $H_{n-1}(A\cap B)$ I will be even more grateful.

  • 3
    $\begingroup$ Think through the proof for cellular or simplicial homology first, where $A$ and $B$ are subcomplexes such that $A\cup B = X$. The reason why you want to do this is you get a short exact sequence of chain complexes $0 \to C_* (A \cap B) \to C_* A \oplus C_* B \to C_* X \to 0$ $\endgroup$ May 14, 2011 at 19:15

1 Answer 1


Let $\mathcal U=\{\operatorname{int}A, \operatorname{int}B\}$, which is an open covering of $X$. According the Proposition 2.21 in Hatcher, the "inclusion" map $\iota:C^{\mathcal U}_\bullet(X)\to C_\bullet(X)$, where $C^{\mathcal U}_\bullet(X)$ is the subcomplex of $C_\bullet(X)$ spanned by the singular simplices whose image is completely contained in one of the elements of $\mathcal U$, induces an isomorphism in homology.

If $\alpha$ is a homology class, represented by a cycle $z\in C_p(X)$, then the previous paragraph implies that there is a $z'\in C^{\mathcal U}_p(X)$ such that $z$ and $\iota(z')$ are homologous. Now $z'$ is a sum of singular simplices which are either in $A$ or in $B$ (or both): let $z'_1$ be the subsum of those in $A$, and let $z'_2$ the subsum of those not in $A$. Then $z'=z'_1+z'_2$ is the decomposition you want.


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