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I am reading through Hatcher and I came across the following statement and am having trouble making sense of it. I am not sure why elements may be written this way. Any help will be appreciated.

By considering the definition of the relative boundary map we see: Elements of $H_n(X, A)$ are represented by relative cycles: $n$ chains $α ∈ C_n(X)$ such that $∂α ∈ C_{n−1} (A).$ A relative cycle $α$ is trivial in $H_n(X, A)$ iff it is a relative boundary: $α = ∂β + γ$ for some $β ∈ C_{n+1} (X)$ and $γ ∈ C_n(A).$

dinky
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2 Answers2

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The relative chains are $C_n(X,A)=C_n(X)/C_n(A)$. Let $\bar\partial$ be the boundary operator on these reduced chains. Then $\bar\partial[\alpha]=0\in C_{n-1}(X)/C_{n-1}(A)$ if and only if $\partial \alpha \in C_{n-1}(A)$. Here I am using $[\alpha]$ to denote the coset represented by $\alpha$.

Similarly, $\bar\partial[\beta]=[\alpha]$ implies that $\partial \beta \in [\alpha]$, which implies $\partial \beta =\alpha+\gamma$ for some $\gamma\in C_n(A)$. (Bear in mind that $[\alpha]=\alpha+ C_n(A)$ as a coset.) It is not hard to see that this implication is actually an "if and only if."

Cheerful Parsnip
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The relative chain groups are defined by $$C_n(X, A) = C_n(X)/C_n(A).$$ The usual boundary map on $C_\ast(X)$ descends to this quotient so that we have a chain complex $(C_\ast(X, A), \partial)$. The relative homology groups of the pair $(X, A)$ are the homology groups of this chain complex.

Now if $\alpha = \partial \beta + \gamma$ for $\beta \in C_{n+1}(X)$ and $\gamma \in C_n(A)$, we have that $[\alpha] = [\partial \beta]$ in the quotient group $C_n(X,A) = C_n(X)/C_n(A)$. Since the boundary map descends to the quotient, we have that $$\tilde{\partial}[\beta] = [\partial \beta],$$ where $\tilde{\partial}: C_{n+1}(X,A) \longrightarrow C_n(X,A)$ is the boundary map on $C_\ast(X,A)$. Therefore $[\alpha]$ is a boundary in $C_n(X,A)$ with respect to the boundary map $\tilde{\partial}$.

Henry T. Horton
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