# Constructing a simplicial map from a diagram

I am trying to read Goerss and Jardine's book (Simplicial Homotopy Theory) and in the proof of Theorem 7.10 (in chapter 1), they claim that there is a simplicial homotopy $$\Delta^n\times\Delta^1\to\Delta^n$$ given by the diagram:

$$\begin{array}{ccccccc} 0&\to &0&\to &\dots&\to &0\\ \downarrow &&\downarrow &&\dots &&\downarrow\\ 0&\to &1&\to&\dots&\to &n \end{array}$$ I am unsure how this actually defines a simplicial map at all. Is there a way to construct a simplicial map from this diagram?

If it helps, the simplicial map is (I think) supposed to be a contraction from the identity to 0.

This simply contracts $$\Delta^n$$ onto the $$0$$-vertex. The diagram you draw encodes the following map $$\mathbf{n} \times\mathbf{1} \to \mathbf{n}$$ where elements $$(m,0)\mapsto 0$$ and $$(m,1) \mapsto m$$. This map induces $$\Delta^n\times \Delta^1 = \operatorname{hom}_\Delta(-, \mathbf{n}) \times \operatorname{hom}_\Delta(-, \mathbf{1}) \cong \operatorname{hom}_\Delta(-,\mathbf{n} \times\mathbf{1}) \to \operatorname{hom}_\Delta(-, \mathbf{n}) = \Delta^n.$$ It could help if you check the effect after geometric realization. This is the standard contracting homotopy of $$|\Delta^n|$$ to $$|\Delta^0| = *$$. Write a formula for this homotopy and compare!
• Perhaps there is a typo in this books’s definition of homotopy, but wouldn’t this homotopy actually be from $0\to\Delta^n$? Certainly the restriction of this homotopy to $\Delta^n\times\{0\}$ is 0, right? At least this is true if the homotopy is defined by $(f,g)\to\varphi(f,g)$, where $\varphi$ is the map $\mathbf{n}\times\mathbf{1}\to\mathbf{n}$. – user722227 Apr 20 '20 at 17:47
Remember that a map between products of simplices is nothing more than a map of the corresponding partially ordered sets. Thus what is given here is the values of the desired map of the vertices of $$\Delta^n\times \Delta^1$$, which uniquely determine the map.