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I want to prove that if $X$ is a path connected space and if $Y$ is nonempty then $$H_0(X,Y)\simeq 0$$

it is sayed that we have this chain: $H_0(Y)\rightarrow H_0(X)\rightarrow H_0(X,Y)\rightarrow 0$ and then is surjetive and then we obtain the result !

but i dont understand why $H_0(Y)\rightarrow H_0(X)$ is surjective (i know that $H_0(X)\simeq \mathbb{Z}$) but i don't understand how we can obtain the result using this surjetivity ?

please help me thank you.

Vrouvrou
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  • What does surjectivity of this map mean in terms of chains? What kind of homology are you using anyway? – Carsten S Nov 09 '13 at 22:07
  • Think about how you showed $H_0(X) = \mathbb Z$. –  Nov 09 '13 at 22:10
  • $H_0(Y)=Z_0(Y)/B_0(Y)=C_0(Y)/B_0(Y)$, and $H_0(X)\simeq \mathbb{Z}$, $i:H_0(Y)\rightarrow \mathbb{Z}$ is surjective means that $i(H_0(Y))=\mathbb{Z}$ – Vrouvrou Nov 09 '13 at 22:14
  • @John to prove that $H_0(X)\simeq \mathbb{Z}$ i proved that $\phi:C_0(X)\rightarrow \mathbb{Z}$ is surjective and i proved that $B_0(X)=\ker \phi $ and i used the factorisation theorem – Vrouvrou Nov 09 '13 at 22:16
  • @CarstenSchultz relative homology – Vrouvrou Nov 09 '13 at 22:19
  • i dont unserstans how to see this – Vrouvrou Nov 09 '13 at 22:20
  • $H_0(X) \cong \mathbb Z [f]$, where $f: {pt} \to X$ is any function as $X$ is path connected. In particular, we can choose $f : {pt} \to Y \subset X$ –  Nov 09 '13 at 22:25
  • I was wondering if this is singular homology or anything else. But if you know already that $H_0(X)\isom\mathbb Z$, then you seem to know all elements of $H_0(X)$. What are they (@John has answered this already)? Are they in the image of $H_0(Y)$? – Carsten S Nov 09 '13 at 22:30
  • @John so we have allwayse $H_0(Y)\simeq \mathbb{Z}$ ? – Vrouvrou Nov 10 '13 at 04:16
  • @CarstenSchultz $H_0(X),H_0(Y)$ is the singular homology and $H_0(X,Y)$ is the relative homology – Vrouvrou Nov 10 '13 at 04:17
  • @John so $H_0(Y)\rightarrow H_0(X)$ is surjective , why this means that $H_0(X,Y)\simeq 0$ ? – Vrouvrou Nov 10 '13 at 04:43
  • $H_0(Y)$ might not be $\mathbb Z$, it depends on the number of path connected component of $Y$, as in that of $X$. The point is that $H_0(Y) \to H_0(X)$ is surjective. Now your sequence is exact, which forces $H_0(X, Y) = 0$. –  Nov 10 '13 at 06:41

1 Answers1

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I think it's useful to know the following fact: $i_*:H_0(A)\to H_0(X)$, induced by the inclusion $i:A\hookrightarrow X$ of subspace $A$, is surjective if and only if $A$ meets each path-component of $X$.

The group $H_0(X)$ is free abelian and a basis is given by the classes $[x_i]$ where $X_i,\ i\in I$, is the set of path components of $X$ and $x_i\in X_i$.
If $A$ meets each path component of $X$, then for each $x_i$ there is an $a_i\in A$ such that $i(a_i)\in X_i$. Since there is a path from $i(a_i)$ to $x_i$, their difference $x_i-i(a_i)$ is a boundary, thus $[x_i]=i_*([a_i])$ and $i_*$ is onto.
Conversely, if $i_*$ is onto, then for each $i\in I$ there is an $a_i\in A$ with $[i(a_i)]=[x_i]$. But this means that $x_i-a_i$ is a boundary. We know that the boundary of a $1$-chain is the sum of the boundaries on the individual path components, where each of these boundaries is a formal sum of points such that the sum of the coefficients $0$. That means that if $x_i$ appears in the boundary on one component, then $a_i$ must cancel it out, so it must lie in the same path component.

Now, $H_0(X,A)$ is zero iff $H_0(X)→H_0(X,A)$ is the zero map iff $H_0(A)→H_0(X)$ is surjective.

Stefan Hamcke
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