Two homotopies relative to the same subspace concatenate after piecewise-linear reparametrisation

The sets $I_0=[0,\tfrac12]$ and $I_1=[\tfrac12,1]$ are closed in $I$: their complements are respectively $I\cap(\tfrac12,\tfrac32)$ and $I\cap(-\tfrac12,\tfrac12)$, traces of open intervals of $\mathbb R$. Hence $D_0=X\times I_0$ and $D_1=X\times I_1$ are closed in $X\times I$, because they are the preimages of $I_0,I_1$ under the continuous time projection.

The maps $a_0:I_0\to I$, $a_0(t)=2t$, and $a_1:I_1\to I$, $a_1(t)=2t-1$, are continuous. Indeed, at any $t_0$ and for any ambient open interval of radius $\varepsilon$ about $a_j(t_0)$, the relative interval of radius $\varepsilon/2$ about $t_0$ maps into it, since $|a_j(t)-a_j(t_0)|=2|t-t_0|$; [L5] turns these intervals into the required subspace neighbourhoods.

Define $r_j:D_j\to X\times I$ by $r_j(x,t)=(x,a_j(t))$. The first component is the restricted product projection and the second is $a_j$ after the time projection, so $r_j$ is continuous by [L2], step 1.2 and [L3].

The maps $K_0:=F\circ r_0:D_0\to Y$ and $K_1:=G\circ r_1:D_1\to Y$ are continuous: for every closed $C\subseteq Y$, $K_j^{-1}[C]=r_j^{-1}[F^{-1}[C]]$ or $r_j^{-1}[G^{-1}[C]]$, which is closed by [L1]. On $D_0\cap D_1=X\times\{\tfrac12\}$ they agree, since $K_0(x,\tfrac12)=F(x,1)=g(x)=G(x,0)=K_1(x,\tfrac12)$.

Thus the displayed clauses define one function $K:X\times I\to Y$. If $C\subseteq Y$ is closed, then $K^{-1}[C]$ is the union of $K_0^{-1}[C]$, regarded as a closed subset of $X\times I$ through the closed subspace $D_0$, and $K_1^{-1}[C]$, regarded likewise through $D_1$. Each is closed by [L2] and steps 1.1 and 3.1, so their union is closed by [L4]. Hence $K$ is continuous by [L1].

At $t=0$ the first clause gives $K(x,0)=F(x,0)=f(x)$, and at $t=1$ the second gives $K(x,1)=G(x,1)=h(x)$. If $a\in A$, both clauses give the common value $f(a)=g(a)=h(a)$ for every $t$. Therefore $K$ is a homotopy from $f$ to $h$ rel $A$.