# Kerodon exercise 1.1.2.8

I'm doing the exercise 1.1.2.8. from Lurie's Kerodon: https://kerodon.net/tag/000T

The link is straight to the exercise does not contain anything else but the exercise itself, so I see no reason to retype it here. However, if someone thinks this should be done, I will do it.

Let $$\Sigma = \{ (\sigma_0,\dots, \sigma_n) \in S^n_{n-1} \mid S_{\partial^{n-1}_j}(\sigma_k) = S_{\partial^{n-1}_{k-1}}(\sigma_j) \text{ for } 0 \leq j < k \leq n \}$$. We have an injection $$\Phi\colon \mathsf{Hom}_{\mathsf{Set}_{\Delta}}(\partial \Delta^n, S) \to \Sigma$$ given by $$f \mapsto (f\circ \Delta^{\partial^n_j})_{0 \leq j \leq n}$$. It is asked to prove that $$\mathrm{im}(\Phi) = \Sigma$$. The implication $$\mathrm{im}(\Phi) \subseteq \Sigma$$ is straightforward. The implication $$\Sigma \subseteq \mathrm{im}(\Phi)$$ is doesn't seem hard as well, but I can't complete a proof due to my lack of experience in combinatorial matters, probably.

We clearly have $$\partial \Delta^n = \bigcup_{0 \leq j \leq n} \Delta^{[n]\setminus\{j\}}$$. There are unique order-preserving bijections $$a_j\colon [n]\setminus\{j\} \to [n-1]$$. Let $$(\sigma_0,\dots, \sigma_n) \in \Sigma$$. Via the Yoneda lemma, we can idenity $$\sigma_i$$ with a map $$\tilde{\sigma}_i\colon \Delta^{n-1}\to S$$. Explicitly, $$(\tilde{\sigma}_i)_m(u) = S_u(\sigma_i)$$ for any $$[m] \in \Delta$$ and any morphism $$u\colon [m]\to [n-1]$$. Thus, we can take the union $$\bigcup \tilde{\sigma}_i$$ to obtain a map $$f\colon \partial \Delta^n\to X$$ with sends $$u\colon [m]\to [n]$$ such that $$\mathrm{im}(f) \subseteq [n]\setminus\{j\}$$ to $$(\tilde{\sigma}_i)(a_j\circ u)$$. My idea (of which I'm almost sure) is that the condition $$S_{\partial^{n-1}_j}(\sigma_k) = S_{\partial^{n-1}_{k-1}}(\sigma_j)$$ for $$j < k$$ should guarantee that the map $$f$$ is well-defined, that is, that it agrees on $$\Delta^{[n]/\{j\}}\cap \Delta^{[n]\setminus\{k\}}$$. Of course, $$u$$ factors through $$[n]\setminus\{j,k\}$$ by assumption.

What can be done next? I've spent some time playing with simplicial operators, but it led me nowhere.

Your solution looks to me like it works, but here is a comment on how you might simplify the combinatorics. Note that we can rephrase the exercise as the statement that we have an equalizer $$\hom_{\mathrm{Set}_{\Delta}}(\partial\Delta^{n}, S_{\bullet}) \to \prod_{0 \leq j \leq n} S_{n-1} \rightrightarrows \prod_{0 \leq j < k \leq n} S_{n-2}$$ where one of the parallel maps acts as $$\vec{\sigma} \mapsto d_{j}(\sigma_{k})$$ and the other as $$\vec{\sigma} \mapsto d_{k-1}(\sigma_{j})$$. This is a rephrasing because the subset of $$\prod_{j} S_{n-1}$$ described in the exercise is the standard construction in $$\mathrm{Set}$$ of the equalizer of these two maps, and the exercise is asking you to show that $$f \mapsto \{f \cdot \delta^{j}\}$$ gives a bijection between $$\hom_{\mathrm{Set}_{\Delta}}(\partial\Delta^{n}, S_{\bullet})$$ and this subset.
Because $$\hom_{\mathrm{Set}_{\Delta}}(-, S_{\bullet})$$ preserves limits, it suffices by Yoneda to show that the preimage diagram $$\coprod_{0 \leq j < k \leq n} \Delta^{[n-2]} \rightrightarrows \coprod_{0 \leq j \leq n} \Delta^{[n-1]} \to \partial\Delta^{n}$$ is a coequalizer in $$\mathrm{Set}_{\Delta}$$. Colimits are created pointwise in $$\mathrm{Set}_{\Delta}$$, so this in turn amounts to showing that $$\coprod_{0 \leq j < k \leq n} \Delta^{[n-2]}_{m} \rightrightarrows \coprod_{0 \leq j \leq n} \Delta^{[n-1]}_{m} \to (\partial\Delta^{n})_{m}$$ is a coequalizer for all $$m$$.
The proof of this last statement is the meat of this exercise, and it's based on the observation in your answer. It's perhaps simplest to show that $$(\partial\Delta^{n})_{m}$$ is in bijection with the standard construction of the coequalizer as the quotient $$Q$$ of $$\coprod_{j}\Delta^{[n-1]}_{m}$$ by the equivalence relation generated by $$(j, \delta^{k-1} \cdot \alpha) \sim (k, \delta^{j} \cdot \alpha)$$ for all $$j < k$$ and $$\alpha : [m] \to [n-2]$$. An element of $$(\partial\Delta^{n})_{m}$$ is a non-surjective map $$\alpha : [m] \to [n]$$. Since it's non-surjective there's some $$j$$ that's not in its image, and so there's some $$\alpha_{j} : [m] \to [n-1]$$ such that $$\alpha = \delta^{j} \cdot \alpha_{j}$$. This means that the map $$Q \to (\partial\Delta^{n})_{m}$$ is surjective. To show that it's injective, consider some pairs $$(j, \alpha)$$ and $$(k, \beta)$$ in $$\coprod_{j} \Delta^{[n-1]}_{m}$$ such that $$\delta^{j} \cdot \alpha = \delta^{k} \cdot \beta$$. If $$j = k$$ then $$\alpha = \beta$$ by the injectivity of $$\delta^{j}$$, and so $$(j, \alpha) = (k,\beta)$$. If $$j \not= k$$ then we can assume without loss of generality that $$j < k$$. Since neither $$j$$ nor $$k$$ lies in the image of $$\delta^{j} \cdot \alpha = \delta^{k} \cdot \beta$$, there's some $$\gamma : [m] \to [n-2]$$ such that $$\delta^{j} \cdot \alpha = \delta^{k} \cdot \beta = \delta^{k} \cdot \delta^{j} \cdot \gamma$$ By the injectivity of $$\delta^{j}$$ and $$\delta^{k}$$ and the fact that $$\delta^{k} \cdot \delta^{j} = \delta^{j} \cdot \delta^{k - 1}$$ we therefore have $$(j, \alpha) = (j, \delta^{k-1} \cdot \gamma) \sim (k, \delta^{j} \cdot \gamma) = (k, \beta)$$ This means that $$Q \to (\partial\Delta^{n})_{m}$$ is injective, hence bijective, giving $$(\delta\Delta^{n})_{m}$$ the universal property of the coequalizer.
We have to note that $${{(a_j)}_{\restriction}}_{[n]\setminus\{j,k\}} = \partial^{n-1}_{k-1}\circ a_{jk}$$ and $${{(a_k)}_{\restriction}}_{[n]\setminus\{j,k\}} = \partial^{n-1}_{j}\circ a_{jk}$$ where $$a_{j,k}$$ is the unique order-preserving bijection $$[n]\setminus\{j,k\} \to [n-2]$$.