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


 * Feel free to post it up - but you are encouraged to set up a username.


 * 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)
 * The above is the solution for the continuous case. However, it might not be $C^\infty$ at $t=\frac 1 2$. Some anonymous person has added a reference for the continuity of the mollifier. It might be worth a proof here on PW; this function is quite standard, especially in the theory of distributions. --Lord_Farin 13:10, 15 October 2011 (CDT)
 * Sounds good. Whatever you can do to enhance it - it's a little over my head till I do some study. --prime mover 07:11, 16 October 2011 (CDT)