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Consider the set of all path homotopy classes of paths in $X$ with $[f]\cdot[g]=[f*g]$ defining a binary operation. We have a groupoid with the following conditions:

1) $[c_p][f]=[f]=[f][c_q]$ where $p=f(0)$, $q=f(1)$ and $[c_r]$ denotes a path constant at $r$.

2) $([f]\cdot[g])\cdot[h] =[f]( [g][h])$ where $f(1)=g(0)$ and $g(1)=h(0)$

3) every path has an inverse.

How can I prove condition 2 (associativity of groupoid) by use of positive linear maps? this proof is in Munkres book but I don't understand it.

Rahmani
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    It's going to be hard to give a better proof than a textbook. The advantage of a site like this is that you can precisely describe at which point you are stuck and what confuses you. Please do so. – Rasmus Nov 17 '13 at 15:17
  • please check my edit. – drhab Nov 17 '13 at 15:28
  • Why We use of positive linear map ? Indeed What work do this linear maps for us to prooving the associativity property? – Rahmani Nov 18 '13 at 06:32
  • Can you prove (1)? This also uses certain linear maps. – Dan Petersen Nov 18 '13 at 13:54
  • To prove (1) : Let C be the constant path at 0 and i be the identity path (i(t)=t) @@@ C(1)=0=i(0) so Ci from I to I be a path and i so .I is contractible so ci is homotopic to i .hence $f$ o ( ci) is homotopic to $f_0$ (i) = c(2t) if 0<t<1/2 and i(2t-1)=2t-1 for 1/2<t<1 .that is equal to f(0)=p=c_p(2t) where 0<t<1/2 and f(2t-1) where 1/2<t<1 that equal to c_p f (t) and so we have fo(i)=f so c_p*f is homotopic to f so the (1) is proved . – Rahmani Nov 18 '13 at 14:08
  • see the https://docs.google.com/file/d/0B7Z2gvey4ZR5S0VxSmtyU1l2Mm8/edit?usp=drive_web&urp=http://www.google.com/url?sa%3Dt%26rct%3Dj%26q%3Dtopology%2520 page 328 the fundamental group – Rahmani Nov 18 '13 at 14:23
  • http://mathoverflow.net/a/80369 – Dustan Levenstein Nov 19 '13 at 15:26
  • If you use paths as maps $f:[0,r]\to X$ where $r \geqslant 0$ then composition of paths is strictly associative and has strict identities, i.e. forms a category. See the book "Topology and Groupoids" (first edition, 1968). – Ronnie Brown Feb 09 '14 at 10:43

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