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I have this question and I'd like an idea to solve it:

If $Z_p(X,Y)=\lbrace \sigma\in C_p(X), \partial\sigma\in C_{p-1}(Y)\rbrace$,

$1)$prove that $H_p(X,Y)$ is isomorphic to $Z_p(X,Y)/(B_p(X)+C_p(Y))$,

$2)$ deduce that $H_0(X,Y)$ is the free module generated by the path connected components of $X$ that do not contain points of $Y$

$Z_p(X,Y)=\ker(\partial_p: C_p(X,Y)\rightarrow C_{p-1}(X,Y))$ $B_p(X,Y)=Im(\partial_{p+1}:C_{p+1}(X,Y)\rightarrow C_p(X,Y))$

$C_p(X,Y)=C_p(X)/C_p(Y)$

$H_p(X,Y)=Z_p(X,Y)/B_p(X,Y)$

Please help me

Thank you.

share|improve this question
    
The definitions of $Z_p(X,Y)$ are not equivalent. Notice that $\operatorname{ker} \partial_p^r = \{ [\sigma]\in C_p(X,Y) \mid \partial \sigma \in C_{p-1}(Y) \}$ is not isomorphic to $\{ \sigma \in C_p(X) \mid \partial \sigma \in C_{p-1}(Y)\}$. Different $\sigma,\sigma'\in C_p(X)$ could have $[\sigma]=[\sigma']$ when they only differ by some $p$-chain in $Y$. –  Christoph Nov 6 '13 at 19:21
    
what i must do please –  Vrouvrou Nov 6 '13 at 19:29
    
Try to understand what I said and figure out why you have to non-equivalent definitions of $Z_p(X,Y)$ and which one should be used. If this is homework, you may cite the exact question. Also please stop just replying with "please help me", show some effort! –  Christoph Nov 6 '13 at 19:32
    
i tryd but i dont find any thing –  Vrouvrou Nov 6 '13 at 21:56
    
it's not a homework, i don't understand what is your method to solve this but i find it in:"INTRODUCTION TO ALGEBRAIC TOPOLOGY AND ALGEBRAIC GEOMETRY" as a proposition –  Vrouvrou Nov 8 '13 at 16:27

2 Answers 2

Hints:

  1. Show that the elements of $Z_0(X,Y)$ are just linear combinations of points of $X$.
  2. Show that $x-y\in B_0(X)$ if and only if there is path from $x$ to $y$.
share|improve this answer
    
Sorry but i stil dont know how to prove that $H_p(X,Y)$ is isomorphic to $Z_p(X,Y)/(B_P(X)+C_P(Y))$ ? –  Vrouvrou Nov 5 '13 at 19:46
    
I thought that was given and you wanted to show that $H_0(X,Y)$ is the free module generated by the path components of $X$ that don't meet $Y$? –  Christoph Nov 5 '13 at 19:57
    
no first i have to prove the isomorphism after that i prove that $H_0(X,Y)$ is generated by... –  Vrouvrou Nov 5 '13 at 20:04
    
can you help me please –  Vrouvrou Nov 5 '13 at 20:10
    
Please edit your question to state what you want to show, what definitions you use, what you have tried, and where you got stuck. –  Christoph Nov 5 '13 at 21:06

Use that relative homology splits over its path components and assume that $X$ has a single path-component. Then for any pair of chains $p\in C_0(X) = Z_0(X,Y)$ and $p^\prime \in C_0(Y)$ we can find an element $s\in Z_1(X)$ such that $\partial s = p-p^\prime$. Hence $[p] = [p^\prime]$ but $[p^\prime] = 0$.

share|improve this answer
    
Update: I also just realised that someone posted a very similar question yesterday: math.stackexchange.com/questions/550616/… –  M.B. Nov 4 '13 at 15:06
    
How to prove the isomorphism ,please ? –  Vrouvrou Nov 5 '13 at 20:58
    
Uhm. Try to do it with a single path component (as above) and then use then page i cite to complete the argument. –  M.B. Nov 6 '13 at 1:26
    
what i must do at first i dont understand ?, supose that $X$ has a singule path component ? –  Vrouvrou Nov 11 '13 at 16:50

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