Questions in the proof of $H_n(X)\simeq H_n(A)\oplus H_n(X, A)$

algebraic-topologyhomology-cohomology

I am working on the following proof and have two questions on it:

Theorem If $A\subset X$ is a retract of $X$ then, $$H_n(X)\simeq H_n(A)\oplus H_n(X, A),$$ all $n\geq 0$.

Proof. Let $r:X\longrightarrow A$ be a retraction. Since $r\circ \imath=id_A$ it follows $r_*\circ \imath_*=id_{H_n(A)}$ hence $\imath_*$ is injective. Consider the exact homology sequence of the pair $(X, A)$: $$\ldots\longrightarrow H_n(A)\stackrel{\imath_*}{\longrightarrow} H_n(X)\stackrel{\jmath_*}{\longrightarrow} H_n(X, A)\stackrel{\partial}{\longrightarrow}H_{n-1}(A)\longrightarrow\ldots$$ Since $\imath_*$ is injective we find $\textrm{ker}(\imath_*)=0=\textrm{im}(\partial)$. But this says $\jmath_*:H_n(X)\longrightarrow H_n(X, A)$ is onto for all $n\geq 0$. In other words, for all $n\geq 0$, we have a short exact sequence $$0\longrightarrow H_n(A)\stackrel{\imath_*}{\longrightarrow} H_n(X)\stackrel{\jmath_*}{\longrightarrow}H_n(X, A)\longrightarrow 0,$$ which splits. Therefore, for all $n\geq 0$, $H_n(X)\simeq H_n(A)\oplus H_n(X, A)$.

My questions:

  1. Why do we have from $r_*\circ \imath_*=id_{H_n(A)}$ that $\imath_*$ is injective? Is it because $id_{H_n(A)}$ is an isomorphismus and $r_*$ is injective? If yes, why is $r_*$ injective?
  2. How exactly do we conclude that $\jmath_*:H_n(X)\longrightarrow H_n(X, A)$ is surjective? I oversee here the keypoint…. What does $\textrm{ker}(\imath_*)=0=\textrm{im}(\partial)$ tells us about $\jmath_*$?

Many thanks for some help!

Best Answer

  1. All you need is that $id_{H_n(A)}$ is injective. Think about what would happen if $i_*$ failed to be injective: could $id_{H_n(A)}$ still be injective?

  2. Since $\partial$ has image $0$, it is the zero map. Therefore all of $H_n(X,A)$ is in the kernel of $\partial$, which by exactness means that all of $H_n(X,A)$ is in the image of $j_*$, or that $j_*$ is surjective.

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