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Let X be a non-empty space and $x_0\in X$, then we have that $H_k(X,x_0)\cong\tilde{H}_k(X)$. (This Statement can be found in 'Algebraic Topology' from Hatcher or slightly different in 'Topology and Geometry' from Bredon; there $x_0$ is replaced with a non-empty acyclic subspace, but the same argument applies there.)

However, when I tried to derive this i got something different, namely that $\tilde{H}_k(X,x_0)\cong\tilde{H}_k(X)$

Let's first consider the long exact sequence of the pair $(X,x_0)$: $$\cdots\rightarrow H_k(x_0)\rightarrow H_k(X)\rightarrow H_k(X,x_0)\rightarrow H_{k-1}(x_0)\rightarrow\cdots$$

Reducing this sequence and using that $\lbrace x_0\rbrace$ is acyclic yields the following: $$\cdots\rightarrow 0\rightarrow \tilde{H}_k(X)\rightarrow \tilde{H}_k(X,x_0)\rightarrow 0\rightarrow\cdots$$ which implies that $\tilde{H}_k(X)\cong\tilde{H}_k(X,x_0)$.

Of course, this is all the same for $k>0$, but if $k=0$ we have that $\tilde{H}_0(X,x_0)\neq H_0(X,x_0)$, so there must be something wrong but I don't see my mistake.

Thanks for any hint.

Aweygan
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PiStrich
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1 Answers1

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You write that $\tilde H_0(X,x_0)\neq H_0(X,x_0)$, but in fact equality always holds. That is;

Prop. For nonempty $A\subset X$, $$\tilde H_k(X,A)\cong H_k(X,A) \qquad \forall k\geq 0.$$

As you have observed, the claim does not follow from the LES when $k=0$. Here the claim must use the definition of reduced homology. See the answer to this question for a proof of the $k=0$ case.

o.h.
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  • Thank you very much. i thought one could derive from the long exact sequence of the relative chain complex that $H_(X,A)\cong\tilde{H}_k(X,A)\oplus\mathbb{Z}$ in the same way one would do it for $H_k(X)$. Your link was very helpful! – PiStrich Feb 14 '19 at 11:38
  • No problem, I remember having the same question when I took algebraic topology :) – o.h. Feb 14 '19 at 11:41