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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 $0<j<n$. 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 \begin{equation} 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}. \end{equation} 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 \begin{equation} z\left\langle k,k+1\right\rangle \in Z_{1}\qquad\text{for each }k\in\left\{ 0,1,\ldots,g-1\right\} . \end{equation} 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 $0<j<n$), 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 \begin{equation} \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}. \end{equation} 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

\begin{equation} 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\} . \end{equation} 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.

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