## 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, $\psi|_X=g\circ f$.