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When you consider the long homology sequence (of spaces $A,X$ , with $A$ subspace of $X$) you need to define an homomorphism from $H_q(X,A)$ to $H_{q-1}(A)$ to obtain the long homology sequence from the short one involving $0 \to H_q(A) \to H_q(X) \to H_q(X,A) \to0$.

The homomorphism you construct from $H_q(X,A) \to H_{q-1}(A)$ is the border homomorphism (it is an easy check that it is well defined in the quotient and its image can be thought in $H_{q-1}(X,A)$).

The main problem I see is that the homomorphism you construct is the zero one (and it does not make sense that you construct it if later is homomorphism zero). I mean, you know that the map from $H_q(A) \to H_q(X)$ is injective and the one from $H_q(X) \to H_q(X,A)$ surjective so by exactness you conclude that the kernel of this ''border'' map is the hole $H_q(X,A)$.

Any idea?

Crostul
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DCao
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    It's not true that the map $H_q(A) \to H_q(X)$ is injective, neither is it true that the map $H_q(X) \to H_q(X, A)$ is surjective. – Balarka Sen May 12 '15 at 21:07
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    In fact the boundary map measures precisely the extent to which both of your claims fail to be true! That's why it's so important. – Qiaochu Yuan May 13 '15 at 17:00

1 Answers1

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The map on the chain level $C_*(A) \to C_*(X)$ might be injective, but the induced map on homology $H_*(A) \to H_*(X)$ need not be. As an example, set $X = \Bbb R^2, A = S^1$. Then the map $\Bbb Z \cong H_1(S^1) \to H_1(\Bbb R^2) \cong 0$ is necessarily zero.

Similarly, the map on homology $H_*(X) \to H_*(X, A)$ need not be surjective. Consider $X = [0,1]$ and $A = \{0,1\}$. Then $H_1(X) \cong 0$ but $H_1(X,A) \cong \Bbb Z \not\cong 0$.

In both of these examples, you can check that the boundary map on homology is nontrivial.

Rolf Hoyer
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