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In this question "Relative homology of a retract: Help to understand a proof." for example. It is stated, among other things, that the border operator that takes a relative cycle from $H_n(X,A)$ to its border in $H_{n-1}(A)$ is actually the $0$ operator. By saying the following sequence is exact: $$0 \to H_n(A) \to H_n(X) \to H_n(X,A) \to 0.$$

But this means that a chain in $X$ that has a border contained in $A$ actually has no border. But why isn't a $\mathbb{R}^2$ and the the open disk a counterexample? Surely the open disk is a retract of $\mathbb{R}^2$, but it is easy to imagine a 1-chain in $\mathbb{R}^2$ that has a border in the disk and its border is not $0$.

(Or the closed disk and a smaller closed disk.)

JKEG
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1 Answers1

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But this means that a chain in $X$ that has a border contained in $A$ actually has no border. But why isn't a $\mathbb{R}^2$ and the the open disk a counterexample? Surely the open disk is a retract of $\mathbb{R}^2$, but it is easy to imagine a 1-chain in $\mathbb{R}^2$ that has a border in the disk and its border is not $0$.

Be careful! To say that element is zero in $H_n(A)$ really means that that element is a $n$-boundary inside $A$. If for example you take a relative $1$-cycle in $H_1(R^2,D^2)$ that is represented by a single arc $c$ whose two endpoints lie inside $D^2$, then under a retract $r:R^2 \to D^2$ this arc becomes an arc $r \circ c$ inside $D^2$ with the same boundary as the original $c$, so $\partial c$ is a $0$-boundary in $A$. Stated differently, we have $[c] \in H_1(R^2,D^2)$, and $$ \partial[c] = [\partial c] = r_*(i_*[\partial c]) = r_*(0) = 0 \in H_0(A),$$ where $i:D^2 \to R^2$ is the inclusion.

Alex Provost
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