From Rotman's Algebraic Topology:
Assume $\phi_0 \simeq \phi_1$ is a free homotopy, where $\phi_i : X \rightarrow Y$ is continuous for $i=0,1$. Choose $x_0 \in X$ and let $\lambda $ denote the path $F(x_0, -)$ in $Y$ from $\phi_0(x_0)$ to $\phi_i(x_0)$. Then there is a commutative diagram $\psi \phi_{1*} = \psi_{0*}$ where $\psi : \pi_1(Y, \phi_1(x_0)) \rightarrow \pi_1(Y, \phi_0(x_0))$ is an isomorphism $[g] \mapsto [\lambda*g * \lambda^{-1}]$ and $\phi_{i*} : \pi_1(X, x_0) \rightarrow \pi_1(Y, \phi_1(x_0))$.
Let $f$ be a closed path in $X$ at $x_0$, define $G : I \times I \rightarrow Y$ by $G(t,s) = F(f(t), s)$.
$G : \phi_0 f \simeq \phi_1 f$. Consider two triangulations of the square $I \times I$:
$T1:$ Draw both diagonals in a square with intersection point $r$. Draw a line from $r$ to the top of the square (point $c$). Draw a line from $r$ to the midpoint between $c$ and the top left of the square (point $d$).
$T2:$ A square with both diagonals drawn in.
Define $H : I \times I \rightarrow I \times I$ as the affine map that:
$(1.)$ Collapses the left(right) edge of the square $T1$ to the single point on the bottom left(right) of the square $T2$.
$(2.)$ Maps the segment $rc$ to the upper right of diagonal segment in $T2$.
$(3.)$ Maps the segment $rd$ to the upper left diagonal segment in $T2$.
$(4.)$ Maps the bottom triangle of $T1$ to the bottom triangle of $T2$.
Define $J = G \circ H : I \times I \rightarrow Y$ as the relative homotopy (via gluing lemma between triangles):
$J : \phi_0 \circ f \simeq (\lambda * (\phi_1 \circ f)) * \lambda^{-1}$ rel $\dot I$
My question is: How is $J$ the appropriate relative homotopy?
I can see that $J(t,0) = G(H(t,0)) = F(f(t), 0) = \phi_0 f (t)$, but I'm having trouble seeing how $J(t,1) = (\lambda * (\phi_1 \circ f)) * \lambda^{-1}$.
Anyone have any ideas?