I am reading kerodon, and I am stuck on exercise 1.1.2.8., which characterizes the simplicial maps of $\partial \Delta^{n}$ on a simplicial set $S_{\cdot}$ as lists $(\sigma_{1},...,\sigma_{n})$ of $n-1$-simplices on $S$ such that $d_{i}(\sigma_{k}) = d_{k-1}(\sigma_{i})$ for $i<k$ by composing such a simplicial map $f$ with the coface maps $\delta^{i}_{n}$ on $\Delta^{n-1}$ induced by the map that skips the ith element on $\Delta$.
They can be seen as $n-1$ simplices because of the fact that $f \circ \delta^{i}_{n}$ is a map from $\Delta^{n-1}$ on $S$, which is $S_{n-1}$ by yoneda lemma. However I fail to see the identites mentioned.
Let $\sigma_{k} = f \circ \delta^{i}_{k}$. Then $d_{i}(\sigma_{k}) = d_{i}(f \circ \delta^{k}_{n}) = f \circ d_{i} \circ \delta^{k}_{n}$. I think I need to some simplicial identity to get the desired identity, but the two maps are not from the same structure. Any help is appreciated.
You need to be a bit careful with the maps you consider, as you noted yourself.
The maps $d_j\colon[n-1]\to[n]$ induce maps $S_n\to S_{n-1}$. If however you identify $n$-simplices of $S$ with maps from $\Delta^n$ to $S$, the face maps become composition with $\delta^j\colon\Delta^{n-1}\to\Delta^n$: $$\begin{align*} (\delta^j)^*\colon\hom(\Delta^n,S)&\to\hom(\Delta^{n-1},S)\\ d_j\colon S_n&\to S_{n-1} \end{align*} $$ (If you now imagine the isomorphisms (from the Yoneda lemma) between the top row and the bottom row you obtain a commutative square.)
Now, since you write an $n-1$ simplex of $S$ as $\sigma_k=f\circ\delta^k\colon\Delta^{n-1}\to\partial\Delta^n\to S$, when you write $d_i(\sigma_k)$ what you mean is the face map as depicted in the top row above, i.e. $d_i(\sigma_k)=\sigma_k\circ\delta^i$. You should now be able to make use of simplicial identities.