Mapping Cone Theorem

Mapping cone
Let $f:(X,x_0)\to (Y,y_0)$ be a map in $\mathscr{PT}$ . We construct the mapping cone $Y\cup_f CX=Y\vee CX/\sim$ , where $[1,x]\in CX$ is identified with $f(x)\in Y$ for all $x\in X$ .

Proposition:
For any map $g: (Y,y_0)\to (Z,z_0)$ we have $g\circ f\simeq z_0$ if and only if $g$ has an extension $h: (Y\cup_f CX,*)\to(Z,z_0)$ to $Y\cup_f CX$ .

Proof:
By an earlier proposition (2.32 in \cite{Switzer2002}), $g\circ f\simeq z_0$ iff $g\circ f$ has an extension $\psi: (CX,*)\to (Z,z_0)$ .

( $\implies$ ) If $g\circ f\simeq z_0$ , define $\tilde{h}: Y\vee CX\to Z$ by $\tilde{h}(y_0,[t,x])=\psi[t,x]$ , $\tilde{h}(y,[0,x])=g(y)$ . Note that $\tilde{h}(y_0,[0,x])=\psi[0,x]=g(y_0)=z_0$ . Since $\displaystyle \tilde{h}(y_0,[1,x])=\psi[1,x]=gf(x)=\tilde{h}(f(x),[0,x]),$ $\tilde{h}$ induces a map $h: Y\cup_f CX\to Z$ which satisfies $h[y,[0,x]]=\tilde{h}(y,[0,x])=g(y)$ . That is $h|_Y=g$ .

( $\impliedby$ ) If $g$ has an extension $h: (Y\cup_f CX,*)\to (Z,z_0)$ , then define $\psi: CX\to Z$ by $\psi([t,x])=h[y_0,[t,x]]$ . We have $\psi([0,x])=h[y_0,[0,x]]=z_0$ . Then $\displaystyle \psi([1,x])=h[y_0,[1,x]]=h[f(x),[0,x]]=gf(x).$ That is, $\psi|_X=g\circ f$ .