# Simplicial set morphisms from an inner horn to a nerve are determined on the spine

This is a subject I'm completely new to, so I am a bit wary of my own proofs.

I will follow the notations of Charles Rezk, Stuff about quasicategories, as of 5 April 2020. I want to prove the following fact, which is an auxiliary observation in his proof of Proposition 5.9:

Lemma 5.9a. Let $$C$$ be a category, and let $$NC$$ be its nerve. Let $$n\in\mathbb{N}$$ and $$j\in\mathbb{N}$$ be such that $$0. Consider the inclusion $$I^{n}\subseteq\Lambda_{j}^{n}$$ of the spine of $$\Delta^{n}$$ into the $$j$$-th inner horn $$\Lambda_{j}^{n}$$. Then, the canonical restriction map $$r:\operatorname*{Hom}\left( \Lambda_{j}^{n},NC\right) \rightarrow \operatorname*{Hom}\left( I^{n},NC\right)$$ is bijective. (Here, $$\operatorname*{Hom}$$ means morphisms of simplicial sets.)

I've so far been scared off Rezk's proof, partly because I don't understand the meaning of $$I^{S}$$ when $$S$$ is not an interval. [EDIT: Charles has explained to me what $$I^S$$ means. Namely, if $$S$$ is a finite nonempty set of integers, then $$I^S$$ is the subcomplex of $$\Delta^S$$ generated by all the $$0$$-simplices and the $$1$$-simplices $$i_0 i_1, i_1 i_2, \ldots, i_{k-1} i_k$$, where $$S$$ is written in the form $$S = \left\{ i_0 < i_1 < \cdots < i_k \right\}$$. Equivalently, if we let $$m \in \mathbb{N}$$ be such that $$S \cong \left[m\right]$$, then $$I^S$$ is the image of $$I^{\left[m\right]}$$ under the canonical isomorphism $$\Delta^{\left[m\right]} \to \Delta^S$$.] But mainly, I wanted to see whether I could prove one of these things in a pedestrian, combinatorial way.

Question. Is the following proof correct?

Proof of Lemma 5.9a. We shall only prove that $$r$$ is injective, since the surjectivity of $$r$$ follows from Proposition 3.19 op. cit..

This is obvious in the case when $$n\leq2$$ (because in this case, we necessarily have $$n=2$$ and $$j=1$$, and thus $$\Lambda_{j}^{n}=\Lambda_{1} ^{2}=I^{2}=I^{n}$$). Thus, we WLOG assume that $$n>2$$. Hence, $$n\geq3$$.

Let $$f$$ and $$f^{\prime}$$ be two elements of $$\operatorname*{Hom}\left( \Lambda_{j}^{n},NC\right)$$ such that $$r\left( f\right) =r\left( f^{\prime}\right)$$. We must prove that $$f=f^{\prime}$$.

The morphisms $$f$$ and $$f^{\prime}$$ are simplicial set morphisms from $$\Lambda_{j}^{n}$$ to $$NC$$, and they are equal when restricted to the subcomplex $$I^{n}$$ of $$\Lambda_{j}^{n}$$ (since $$r\left( f\right) =r\left( f^{\prime}\right)$$).

Let $$Z$$ be the equalizer of $$f$$ and $$f^{\prime}$$ in the category of simplicial sets. This is simply the subcomplex of $$\Lambda_{j}^{n}$$ given by $$$$Z_{k}=\left\{ z\in\left( \Lambda_{j}^{n}\right) _{k}\ \mid\ f\left( z\right) =f^{\prime}\left( z\right) \right\} \qquad\text{for each } k\in\mathbb{N}.$$$$ Then, $$I^{n}\subseteq Z$$ (since $$f$$ and $$f^{\prime}$$ are equal when restricted to the subcomplex $$I^{n}$$). Thus, $$\left( I^{n}\right) _{0}\subseteq Z_{0}$$ and $$\left( I^{n}\right) _{1}\subseteq Z_{1}$$.

If $$S$$ is a simplicial set, then I shall use the notation "$$s\in S$$" (or "$$s$$ belongs to $$S$$") as a shorthand for "$$s\in S_{k}$$ for some $$k\in\mathbb{N}$$".

Claim 1: Let $$g\in\mathbb{N}$$ and $$z\in\left( \Lambda_{j}^{n}\right) _{g}$$. Assume that $$$$z\left\langle k,k+1\right\rangle \in Z_{1}\qquad\text{for each }k\in\left\{ 0,1,\ldots,g-1\right\} .$$$$ Then, $$z\in Z_{g}$$.

Proof of Claim 1: If $$g=0$$, then this follows immediately from $$z\in\left( \Lambda_{j}^{n}\right) _{g}=\left( \Lambda_{j}^{n}\right) _{0}=\left( I^{n}\right) _{0}\subseteq Z_{0}=Z_{g}$$. Thus, WLOG assume that $$g\neq0$$.

Let $$k\in\left\{ 0,1,\ldots,g-1\right\}$$. Then, $$z\left\langle k,k+1\right\rangle \in Z_{1}$$ (by assumption). In other words, $$f\left( z\left\langle k,k+1\right\rangle \right) =f^{\prime}\left( z\left\langle k,k+1\right\rangle \right)$$ (by the definition of $$Z$$). But $$f\left( z\left\langle k,k+1\right\rangle \right) =f\left( z\right) \left\langle k,k+1\right\rangle$$ (since $$f$$ is a morphism of simplicial sets) and similarly $$f^{\prime}\left( z\left\langle k,k+1\right\rangle \right) =f^{\prime}\left( z\right) \left\langle k,k+1\right\rangle$$. Thus, $$f\left( z\right) \left\langle k,k+1\right\rangle =f\left( z\left\langle k,k+1\right\rangle \right) =f^{\prime}\left( z\left\langle k,k+1\right\rangle \right) =f^{\prime}\left( z\right) \left\langle k,k+1\right\rangle$$.

Now, forget that we fixed $$k$$. We thus know that each $$k\in\left\{ 0,1,\ldots,g-1\right\}$$ satisfies $$f\left( z\right) \left\langle k,k+1\right\rangle =f^{\prime}\left( z\right) \left\langle k,k+1\right\rangle$$. But a $$g$$-simplex $$w$$ in $$NC$$ is uniquely determined by its edges $$w\left\langle 0,1\right\rangle ,w\left\langle 1,2\right\rangle ,\ldots,w\left\langle g-1,g\right\rangle$$ (since it just "consists" of objects and morphisms of $$C$$, and these latter morphisms are precisely $$w\left\langle 0,1\right\rangle ,w\left\langle 1,2\right\rangle ,\ldots ,w\left\langle g-1,g\right\rangle$$, whereas the objects are uniquely determined by the morphisms (here we use $$g\neq0$$)). In other words, if $$w$$ and $$w^{\prime}$$ are two $$g$$-simplices in $$NC$$ such that each $$k\in\left\{ 0,1,\ldots,g-1\right\}$$ satisfies $$w\left\langle k,k+1\right\rangle =w^{\prime}\left\langle k,k+1\right\rangle$$, then $$w=w^{\prime}$$. Applying this to $$w=f\left( z\right)$$ and $$w^{\prime}=f^{\prime}\left( z\right)$$, we conclude that $$f\left( z\right) =f^{\prime}\left( z\right)$$. In other words, $$z\in Z_{g}$$ (by the definition of $$Z$$). This proves Claim 1.

Next, let us recall that the elements of $$\left( \Lambda_{j}^{n}\right) _{g}$$ for a given $$g\in\mathbb{N}$$ are the simplicial operators $$\left[ g\right] \rightarrow\left[ n\right]$$ whose image does not contain $$\left[ n\right] \setminus\left\{ j\right\}$$ as a subset (where, as usual, $$\left[ n\right] =\left\{ 0,1,\ldots,n\right\}$$). In other words, they are the simplicial operators $$\left[ g\right] \rightarrow\left[ n\right]$$ which miss at least one element distinct from $$j$$.

Claim 2: Let $$p$$ and $$q$$ be integers satisfying $$0\leq p\leq q\leq n$$ and $$\left( p,q\right) \neq\left( 0,n\right)$$. Then, the morphism $$\left\langle p,p+1,\ldots,q\right\rangle :\left[ q-p\right] \rightarrow \left[ n\right]$$ belongs to $$Z$$.

Proof of Claim 2: The image of the morphism $$\left\langle p,p+1,\ldots ,q\right\rangle$$ does not contain $$\left[ n\right] \setminus\left\{ j\right\}$$ as a subset (since otherwise, it would contain both $$0$$ and $$n$$ (since $$0), but this would contradict $$\left( p,q\right) \neq\left( 0,n\right)$$). Hence, $$\left\langle p,p+1,\ldots,q\right\rangle \in\left( \Lambda_{j}^{n}\right) _{q-p}$$. For each $$k\in\left\{ 0,1,\ldots ,q-p-1\right\}$$, we have $$$$\left\langle p,p+1,\ldots,q\right\rangle \left\langle k,k+1\right\rangle =\left\langle p+k,p+k+1\right\rangle \in\left( I^{n}\right) _{1}\subseteq Z_{1}.$$$$ Hence, Claim 1 (applied to $$g=q-p$$ and $$z=\left\langle p,p+1,\ldots ,q\right\rangle$$) shows that $$\left\langle p,p+1,\ldots,q\right\rangle \in Z_{q-p}$$. This proves Claim 2.

Claim 3: Let $$p$$ and $$q$$ be integers satisfying $$0\leq p\leq q\leq n$$ and $$\left( p,q\right) \neq\left( 0,n\right)$$. Then, the morphism $$\left\langle p,q\right\rangle :\left[ 1\right] \rightarrow\left[ n\right]$$ belongs to $$Z$$.

Proof of Claim 3: Claim 2 yields that $$\left\langle p,p+1,\ldots ,q\right\rangle$$ belongs to $$Z$$. Thus, $$\left\langle p,p+1,\ldots ,q\right\rangle \left\langle 0,q-p\right\rangle$$ also belongs to $$Z$$ (since $$Z$$ is a simplicial set). In view of $$\left\langle p,p+1,\ldots,q\right\rangle \left\langle 0,q-p\right\rangle =\left\langle p,q\right\rangle$$, this rewrites as follows: The morphism $$\left\langle p,q\right\rangle :\left[ 1\right] \rightarrow\left[ n\right]$$ belongs to $$Z$$. This proves Claim 3.

Claim 4: Let $$p$$ and $$q$$ be integers satisfying $$0\leq p\leq q\leq n$$. Then, the morphism $$\left\langle p,q\right\rangle :\left[ 1\right] \rightarrow \left[ n\right]$$ belongs to $$Z$$.

Proof of Claim 4: If $$\left( p,q\right) \neq\left( 0,n\right)$$, then this follows from Claim 3. Hence, we WLOG assume that $$\left( p,q\right) =\left( 0,n\right)$$. Thus, $$\left\langle p,q\right\rangle =\left\langle 0,n\right\rangle$$.

We have $$n\geq3$$. Hence, the image of the morphism $$\left\langle 0,j,n\right\rangle :\left[ 2\right] \rightarrow\left[ n\right]$$ does not contain $$\left[ n\right] \setminus\left\{ j\right\}$$ as a subset (since it contains only $$2$$ elements of $$\left[ n\right] \setminus\left\{ j\right\}$$). Hence, $$\left\langle 0,j,n\right\rangle \in\left( \Lambda _{j}^{n}\right) _{2}$$. Moreover, $$\left\langle 0,j,n\right\rangle \left\langle 0,1\right\rangle =\left\langle 0,j\right\rangle$$ belongs to $$Z$$ (by Claim 2, applied to $$\left( p,q\right) =\left( 0,j\right)$$), so that $$\left\langle 0,j,n\right\rangle \left\langle 0,1\right\rangle \in Z_{1}$$. Similarly, $$\left\langle 0,j,n\right\rangle \left\langle 1,2\right\rangle \in Z_{1}$$. Combining these two results, we conclude that $$\left\langle 0,j,n\right\rangle \left\langle k,k+1\right\rangle \in Z_{1}$$ for each $$k\in\left\{ 0,1\right\}$$. Thus, Claim 1 (applied to $$g=2$$ and $$z=\left\langle 0,j,n\right\rangle$$) shows that $$\left\langle 0,j,n\right\rangle \in Z_{2}$$. Hence, $$\left\langle 0,j,n\right\rangle \left\langle 0,2\right\rangle \in Z_{1}$$ (since $$Z$$ is a subcomplex). In view of $$\left\langle 0,j,n\right\rangle \left\langle 0,2\right\rangle =\left\langle 0,n\right\rangle =\left\langle p,q\right\rangle$$, this rewrites as $$\left\langle p,q\right\rangle \in Z_{1}$$. This proves Claim 4.

Claim 5: We have $$f\left( z\right) =f^{\prime}\left( z\right)$$ for each $$z\in\Lambda_{j}^{n}$$.

Proof of Claim 5: Fix $$z\in\Lambda_{j}^{n}$$. Thus, $$z\in\left( \Lambda _{j}^{n}\right) _{g}$$ for some $$g\in\mathbb{N}$$. Consider this $$g$$.

Claim 4 shows that $$\left( \Lambda_{j}^{n}\right) _{1}\subseteq Z_{1}$$. But $$z\in\Lambda_{j}^{n}$$. Since $$\Lambda_{j}^{n}$$ is a simplicial set, we thus have

$$$$z\left\langle k,k+1\right\rangle \in\left( \Lambda_{j}^{n}\right) _{1}\subseteq Z_{1}\qquad\text{for each }k\in\left\{ 0,1,\ldots,g-1\right\} .$$$$ Thus, $$z\in Z_{g}$$ (by Claim 1). In view of the definition of $$Z$$, this yields $$f\left( z\right) =f^{\prime}\left( z\right)$$. This proves Claim 5.

Clearly, Claim 5 proves that $$f=f^{\prime}$$. This proves Lemma 5.9a.