Suppose $F_0,F_1: X \rightarrow Y$ and $G_0,G_1: Y \rightarrow Z$ are continuous maps. Suppose that $F_0 \simeq F_1$ and $G_0 \simeq G_1$, show then that $G_0 \circ F_0 \simeq G_1 \circ F_1$.
First off I know there exists continuous maps
$$H_F: X \times [0,1] \rightarrow Y$$
defined via
$$H_F(x,0):=F_0(x), H_F(x,1):=F_1(x)$$
and a continuous map $H_G$
$$H_G: Y \times [0,1] \rightarrow Z$$
such that
$$H_G(y,0):=G_0(y),H_G(y,1):=G_1(y)$$
My question is do I take the piecewise function defined as a new function $H$ as $H_F:=(s,2t)$ for $t \in [0,\frac{1}{2}]$ and define it as $H_G:=(s,2t-1)$ for $t \in (\frac{1}{2},1]$ Then $H$ is defined on
$$H: X \times [0,1] \rightarrow Z$$
And clearly composition of continuous functions is continuous thus $G_0 \circ F_0$,$G_1 \circ F_1$ which map $X$ into $Z$ are both continuous.
My issue is that $H_G$ is defined on $Y \times [0,1]$. Thus my $H$ I'm not sure is defined on $X \times [0,1]$. Am I crazy or does my new homotopy need to just go from $X \times [0,1]$ into $Z$? Also, $H_F(s,t)$ lands in $Y$ not $Z$. Where am I going wrong ?
Best Answer
It just needs the slightest fix: define $H$ as $$H(x,t) = \begin{cases} G_0(H_F(x,2t)) &\quad\text{if $0 \le t \le 1/2$} \\ H_G(F_1(x),2t-1) &\quad\text{if $1/2 \le t \le 1$} \end{cases} $$