How do we differentiate homotopies?

Question

The problem setting is the following: we have two functions, $f:[0,1]\rightarrow[0,1]$ and $g:[0,1]\rightarrow[0,1]$, $0\leq\lambda\leq 1$, $f^{\prime}(t) > 0$, $g^{\prime}(t) > 0$, $f(0) = 0$, $f(1) = g(1) = 1$ and $f^{\prime}(t)\geq g^{\prime}(t)$. Then consider a homotopy between $f$ and $g$: $H:[0,1]\times\Lambda\rightarrow[0,1]$ such that $H(t,0) = g(t)$ and $H(t,1) = f(t)$. My question is: how do we differentiate $\lambda H$ with respect to $\lambda$ and $t$?

Answer 1

Let $\mathcal H(\lambda, t) = \lambda H(t,\lambda)$. Then \begin{align*} \frac{\partial\mathcal H}{\partial \lambda} &= H(t,\lambda) + \lambda\frac{\partial H}{\partial\lambda} \qquad\text{and} \\ \frac{\partial\mathcal H}{\partial t} & = \lambda\frac{\partial H}{\partial t}.\end{align*} If we don't know the homotopy, that's all we can say. If you took the straight-line homotopy between the functions, we'd have $H(t,\lambda) = \lambda f(t)+(1-\lambda)g(t)$, and then we could say $\dfrac{\partial H}{\partial \lambda} = \lambda f'(t)+(1-\lambda)g'(t)$, and then you obtain $\dfrac{\partial\mathcal H}{\partial\lambda}$ by substitution.