[tex]

f(t,0)=x_1(t),\quad f(t,1)=x_2(t),\quad f(0,s) = x_1(0) = x_2(0),\quad f(1,s) = x_1(1) = x_2(1)

[/tex]

How do you prove, that there also exists a homotopy [itex]g:[0,1]^2\to G[/itex] between the two paths so, that for all [itex]s\in [0,1][/itex], the mapping

[tex]

t\mapsto g(t,s)

[/tex]

is differentiable?

This seems a clear claim, but doesn't seem to come easily from the definitions.