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i have that $H_p(X,Y)$ is isomorphic to $Z_p(X,Y)/(B_p(X)+C_p(Y))$, where $Z_p(X,Y)=\lbrace \sigma\in C_p(X), \partial\sigma\in C_{p-1}(Y)\rbrace$

and i want to deduce that $H_0(X,Y)$ is the free module generated by the path connected components of $X$ that do not contain points of $Y$

but i don't know how to prove it ?

can someone help me ?

Thank you

Vrouvrou
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  • What have you tried? (You've been around long enough now that you should know you need to show us what you've tried). – Dan Rust Nov 10 '13 at 18:44
  • @DanielRust the problem is that i don't know what i have to prove exacltly , i know that $Z_0(X,Y)=C_0(X,Y)$ so the elements of $Z_0(X,Y)$ are just linear combinations of points of X but i don't know what i must do after that – Vrouvrou Nov 10 '13 at 18:59
  • Repost: http://math.stackexchange.com/questions/551575/question-on-relative-homology/551619#551619 – M.B. Nov 10 '13 at 23:58

1 Answers1

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Look at the sequence $H_0(Y)\stackrel{i_*}→H_0(X)\stackrel{q}→H_0(X,Y)→0$ which is part of the long exact sequence for a pair. $H_0(X)$ is freely generated by the $[x_i]$ where $X_i, i∈I$, is the set of path components of $X$ and $x_i\in X_i$. The map $q$ is surjective, so $H_0(X,Y)\approx H_0(X)/i_*[H_0(Y)]$. But the image $i_*[H_0(Y)]$ is generated by the $[i(y_j)]$ where $Y_j,\ j\in J$, is the set of path components of $Y$ and $y_j\in Y_j$. Now, $[y_j]=[x_{c(j)}]$ where $c:J→I$ is a function such that $X_{c(j)}$ contains $Y_j$. That means that $$\frac{H_0(X)}{i_*[H_0(Y)]}= \frac{⊕_{i∈I}\Bbb Z⟨[x_i]⟩}{⊕_{c(j),j∈J}\Bbb Z⟨[x_{i(j)}]⟩}= ⊕_{k∈I-c[J]}\Bbb Z⟨[x_k]⟩$$

Stefan Hamcke
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