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.

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