I was wondering if somebody could point me to a resource which proves the fact that two functions being smoothly homotopic is an equivalence relation. It is easy when the homotopy $H:X\times [0,1] \rightarrow Y$ between two functions $f:X\rightarrow Y$ and $g:X\rightarrow Y$ need only be continuous, but seems less than straightforward when it is required to be smooth as well.
The hard part is of course transitivity. I found one reference here: https://proofwiki.org/wiki/Smooth_Homotopy_is_an_Equivalence_Relation , but this seems to require some assumptions on the range space $Y$ that are not clearly stated. Specifically, it defines two homotopies
$A(x,t)= \phi(t) g(x) + (1-\phi(t))f(x)$ and
$B(x,t)= \phi(t) h(x) + (1-\phi(t))g(x)$,
but if we take $t=\frac{1}{2}$ in either, then we get, for example, $A(x,\frac{1}{2})= f(x)+\phi(\frac{1}{2})(g(x)-f(x))$, which does not need to necessarily be in the range space $Y$ unless it $Y$ is assumed convex.
Any help would be greatly appreciated here. Thanks!
The usual trick is to replace the homotopy $H$ by a homotopy which is "constant" for $t$ near zero and $t$ near $1$. Then two such homotopies can be spliced without losing smoothness at the join.
We assume $X$ and $Y$ are smooth manifolds.
To be more precise we want to find a smooth homotopy $H':X\times I\to Y$ with the property that $H'(x,t)=H'(x,0)$ for $t\in[0,1/3]$ and $H'(x,t)=H'(x,1)$ for $t\in[2/3,1]$. To do this define $$H'(x,t)=H(x,\phi(t))$$ where $H$ is your original homotopy, $\phi$ is a smooth function $\Bbb R\to\Bbb R$ with $\phi\equiv0$ on $[0,1/3]$, $\phi\equiv1$ on $[2/3,1]$ and $\phi(t)\in[0,1]$ in all cases. One can construct such a function by taking the indefinite integral of a suitable smooth "bump function".