Kerodon exercise 1.1.2.8

Question

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.

Answer 1

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.

Answer 2

I think I've got it.

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 $[n]\setminus\{j,k\} \to [n-2]$.

I hope this settles this. Any comments are welcome.