Talk:Relative Homotopy is Equivalence Relation

There is an easier way to show transitivity. f~g,g~h. H_{f,g}:XxI->Y the homotopy between f,g and define in the same way H_{g,h}. Then { H_{f,g}(x,2t) t in [0,1/2] H_{f,g}={ { H_{g,h}(x,2t-1) t in [1/2,1]

is an homotopy. -- Special:Contributions/201.252.194.119


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 * I have $\LaTeX$ed the above - I think it goes:


 * $f \sim g, g \sim h: H_{f,g} : X \times I \to Y$ the homotopy between $f,g$ and define in the same way $H_{g,h}$. Then:
 * $H_{f, g} = \begin{cases} H_{f,g}(x,2t): t \in [0,1/2] \\ H_{g,h}(x,2t-1): t \in [1/2,1]\end{cases}$


 * Also it needs to be considerably expanded and explained. --prime mover 02:23, 8 October 2011 (CDT)