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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.

Vrouvrou
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  • 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 06 '13 at 19:21
  • what i must do please – Vrouvrou Nov 06 '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 06 '13 at 19:32
  • i tryd but i dont find any thing – Vrouvrou Nov 06 '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 08 '13 at 16:27
  • i you can explain me more what is your method i will don't say no – Vrouvrou Nov 08 '13 at 20:50

2 Answers2

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 $y$.
Christoph
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  • 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 05 '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 05 '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 05 '13 at 20:04
  • can you help me please – Vrouvrou Nov 05 '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 05 '13 at 21:06
  • i don't have an idea to prove the isomorphism – Vrouvrou Nov 05 '13 at 21:13
  • Whats the definition of $H_p(X,Y)$? – Christoph Nov 05 '13 at 21:59
  • $H_p(X,Y)=Z_p(X,Y)/B_p(X,Y)$] – Vrouvrou Nov 06 '13 at 05:39
  • can you help me please , i really don't know how to prove the isomorphisme – Vrouvrou Nov 06 '13 at 18:47
  • In your question, write down the definitions of $Z_p(X,Y)$, $B_p(X,Y)$ and $H_p(X,Y)$. Then state exactly what you want to prove, what your ideas are so far and where you're stuck. We can't help you If we don't know what assumptions you're starting off with and what you've tried and where you don't understand something. – Christoph Nov 06 '13 at 18:54
  • hi, so i agree with you in one $Z_0(X,Y)$ are just linear combinations of points of $X$ but i don't understand 2) and how to make the conclusion ? – Vrouvrou Nov 09 '13 at 20:36
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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 $[p^\prime] = 0$.

Community
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M.B.
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  • Update: I also just realised that someone posted a very similar question yesterday: http://math.stackexchange.com/questions/550616/h-0x-a-0-iff-a-meets-each-path-component-of-x – M.B. Nov 04 '13 at 15:06
  • How to prove the isomorphism ,please ? – Vrouvrou Nov 05 '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 06 '13 at 01:26
  • what i must do at first i dont understand ?, supose that $X$ has a singule path component ? – Vrouvrou Nov 11 '13 at 16:50