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!