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I'd like to prove that $\tilde{H}_n(X \cup CA) \simeq H_n(X,A)$, where $CA$ denotes the cone of $A$, i.e $CA = A \times I / A \times \{0\}$ where $A$ is a subset of $X$ topological space.

The first step in order to do so is to prove the following isomorphism $\tilde{H}_n(X \cup CA) \simeq H_n(X \cup CA,CA)$. Since $CA$ is contractible I thought of taking the long exact sequence of the pair and i get $$\longrightarrow H_n(CA) \longrightarrow H_n(X \cup CA) \longrightarrow H_n(X \cup CA,CA) \longrightarrow H_{n-1}(CA)\longrightarrow$$

So if $n-1 \ne 0$ i.e $n > 1$ since $H_i(CA) = 0$ we have the $\tilde{H}_n(X \cup CA) \simeq H_n(X \cup CA,CA)$ since for those $n, \hspace{0.1cm} \tilde{H}_n(X \cup CA) \simeq H_n(X \cup CA)$.

Here is where I got stuck, since the terms are no longer $0$ I don't know how to continue in the case $n=0,n=1$. Any help would be appreciated.

Henno Brandsma
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jacopoburelli
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1 Answers1

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The long exact sequence has two variants: One for unreduced homology groups ands one for reduced homology groups. The latter is $$\dots \to \tilde H_n(A) \to \tilde H_n(X) \to H_n(X,A) \to \tilde H_{n-1}(A) \to \dots $$ Note that $\tilde H_n(Y) = H_n(Y)$ for $n > 0$. See see Hatcher's "Algebraic Topology" p.118 and Suppose that $X$ is a topological space and $x_0\in X$. Prove that $\widetilde{H_n}(X)=H_n(X,x_0)$ for all $n\geq 0$..

Now look at $$H_1(CA) \to H_1(X \cup CA) \to H_1(X \cup CA, CA) \to \tilde H_0(CA) \to \tilde H_0(X \cup CA) \to H_0(X \cup CA,CA) \to 0 $$ Since $H_1(CA) = \tilde H_0(CA) = 0$, you get the desired result.

Paul Frost
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  • Why did you specify that $H_1(CA) = 0$ ? Didn't the case $n=0$ follows from the fact that the sequence is exact and the terms stay between 2 zeros ? – jacopoburelli Mar 21 '21 at 14:46
  • I also don't see where did you use that $\tilde{H}_n(X) \simeq H_n(X,x_0)$ why cone being contractible isn't enough to deduce the isomorphism ? – jacopoburelli Mar 21 '21 at 14:50
  • @jacopoburelli 1. For $n = 0$ yes, but you also asked for $n = 1$ and for that case we need $H_1(CA) = 0$. 2. I didn't use $\tilde H_n(X) \approx H_n(X,x_0)$. Indeed this is completely irrelevant here. The purpose of the link was to give information about the reduced homogy groups and their long exact sequence. See my answer to the linked question. – Paul Frost Mar 21 '21 at 18:02