ScalingStacks

Proof. We prove the proposition by producing particular choices of compositions which give equal (not just homotopic) results.

To construct $h\circ(g\circ f)$ consider the diagram

Note that the composite of the vertical maps in the left-hand column is $\varphi_{3}$. Any choice of $k\in\map(w,x,y,z)$ such that $\varphi_{3}(k)=(h,g,f)$ determines compositions $d_{3}k\in\map(w,x,y)$ and $d_{1}k\in\map(w,y,z)$ with results $g\circ f$ and $h\circ(g\circ f)$ respectively. By considering an analogous diagram we see that such a $k$ also determines compositions $d_{0}k\in\map(x,y,z)$ and $d_{2}k\in\map(w,x,z)$ with results $h\circ g$ and $(h\circ g)\circ f$ respectively, and that for this choice of compositions there is an equality $h\circ(g\circ f)=(h\circ g)\circ f$ of results, as desired.

To show that $f\circ\id_{w}\sim f$ for $f\in\map(w,x)$, let $k=s_{0}(f)\in\map(w,w,x)$. Then $\varphi_{2}(k)=(f,\id_{w})$ and $d_{1}(k)=f$, showing that $f\circ\id_{w}=f$. The proof that $\id_{z}\circ f\sim f$ is similar. ∎