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For an exact sequence $A\rightarrow B\rightarrow C\rightarrow D\rightarrow E$ show that $C=0$ iff the map $A\rightarrow B$ is surjective and $D\rightarrow E$ is injective.

For $(C=0)$ implies ($A\rightarrow B$ is surjective and $D\rightarrow E$ is injective), ok because $Im(A\rightarrow B)=Ker(B\rightarrow 0)=B$ hence $A\rightarrow B$ is surjective and $Ker(D\rightarrow E)=Im(0\rightarrow D)=0$ hence $D\rightarrow E$ is injective.

I don't know for ($A\rightarrow B$ is surjective and $D\rightarrow E$ is injective) implies ($C=0$).

If I show this, I can conclude that $H_{n}(X,A)=0$ iff $H_{n}(X)\simeq H_{n}(A)$. Thank you !

Fundamentalistic
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2 Answers2

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$A \rightarrow B$ surjective implies $\mathrm{im}(A \rightarrow B) = B$ implies $\ker(B \rightarrow C) = B$ implies $\mathrm{im}(B \rightarrow C) = 0$ implies $\ker(C \rightarrow D) = 0$.

However $D \rightarrow E$ injective implies $\ker(D \rightarrow E) = 0$ implies $\mathrm{im}(C \rightarrow D) = 0$ implies $\ker(C \rightarrow D) = C$.

Thus $C = 0$.

aes
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  • $\operatorname{img} (B → C) = 0$, because $\operatorname{ker} (B → C) = \operatorname{img} (A → B) = B$.

  • $\operatorname{img} (B → C) = \operatorname{ker} (C → D)$ by exactness.

Therefore $C \cong \operatorname{img} (C → D) = \operatorname{ker}(D → E) = …$

k.stm
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