### 3.1.2 Anodyne Morphisms

By definition, a morphism of simplicial sets $f: X \rightarrow S$ is a Kan fibration if it is weakly right orthogonal to every horn inclusion $\Lambda ^{n}_{i} \hookrightarrow \Delta ^ n$ for $0 \leq i \leq n$ and $n > 0$ (Definition 1.4.4.3). If this condition is satisfied, then $f$ is weakly right orthogonal to a much larger collection of morphisms.

Definition 3.1.2.1 (Anodyne Morphisms). Let $T$ be the smallest collection of morphisms in the category $\operatorname{Set_{\Delta }}$ with the following properties:

For each $n > 0$ and each $0 \leq i \leq n$, the horn inclusion $\Lambda ^{n}_{i} \hookrightarrow \Delta ^ n$ belongs to $T$.

The collection $T$ is weakly saturated (Definition 1.4.4.12). That is, $T$ is closed under pushouts, retracts, and transfinite composition.

We say that a morphism of simplicial sets $i: A_{} \rightarrow B_{}$ is *anodyne* if it belongs to the collection $T$.

Example 3.1.2.4. Let $i: A_{} \hookrightarrow B_{}$ be an inner anodyne morphism of simplicial sets (Definition 1.4.6.4). Then $i$ is anodyne. The converse is false in general. For example, the horn inclusions $\Lambda ^{n}_0 \hookrightarrow \Delta ^ n$ and $\Lambda ^{n}_{n} \hookrightarrow \Delta ^ n$ are anodyne (for $n > 0$), but are not inner anodyne.

Example 3.1.2.5. For $0 \leq i \leq n$, the inclusion map $\{ i\} \hookrightarrow \Delta ^ n$ is anodyne. To prove this, let $\operatorname{Spine}[n]$ denote the spine of the $n$-simplex, so that the inclusion $\operatorname{Spine}[n] \hookrightarrow \Delta ^ n$ is inner anodyne (Example 1.4.7.7) and therefore anodyne (Example 3.1.2.4). It will therefore suffice to show that the inclusion $\{ i\} \hookrightarrow \operatorname{Spine}[n]$ is anodyne, which is clear (it can be written as a composition of pushouts of the inclusions $\{ 0\} \hookrightarrow \Delta ^1$ and $\{ 1\} \hookrightarrow \Delta ^1$).

We will need the following stability properties for the class of anodyne morphisms:

Proposition 3.1.2.8. Let $f: A_{} \hookrightarrow B_{}$ and $f': A'_{} \hookrightarrow B'_{}$ be monomorphisms of simplicial sets. If either $f$ or $f'$ is anodyne, then the induced map

\[ (A_{} \times B'_{} ) \coprod _{ A_{} \times A'_{} } ( B_{} \times A'_{} ) \hookrightarrow B_{} \times B'_{} \]

is anodyne.

The proof of Proposition 3.1.2.8 will require some preliminaries.

Lemma 3.1.2.9. For every pair of integers $0 < i \leq n$, the horn inclusion $f_0: \Lambda ^{n}_{i} \hookrightarrow \Delta ^ n$ is a retract of the inclusion map $f: (\Delta ^1 \times \Lambda ^ n_ i) \coprod _{ \{ 1\} \times \Lambda ^{n}_ i} ( \{ 1\} \times \Delta ^ n) \hookrightarrow \Delta ^1 \times \Delta ^ n$.

**Proof.**
Let $A_{}$ denote the simplicial subset of $\Delta ^1 \times \Delta ^ n$ given by the union of $\Delta ^1 \times \Lambda ^ n_ i$ with $\{ 1 \} \times \Delta ^ n$. To prove Lemma 3.1.2.9, it will suffice to show that there exists a commutative diagram of simplicial sets

\[ \xymatrix@R =50pt@C=50pt{ \{ 0\} \times \Lambda ^{n}_{i} \ar [r] \ar [d]^{f_0} & A_{} \ar [r] \ar [d]^{f} & \Lambda ^{n}_{i} \ar [d]^{f_0} \\ \{ 0\} \times \Delta ^ n \ar [r] & \Delta ^1 \times \Delta ^ n \ar [r]^-{r} & \Delta ^ n } \]

where the left horizontal maps are given by inclusion and the horizontal compositions are the identity maps. To achieve this, it suffices to choose $r$ to be given on vertices by the map of partially ordered sets

\[ r: [1] \times [n] \rightarrow [n] \quad \quad r(j,k) = \begin{cases} i & \text{ if $j=1$ and $k \leq i$ } \\ k & \text{ otherwise. } \end{cases} \]

$\square$

Lemma 3.1.2.10. Let $X$ be a simplicial set which is the union of a simplicial subset $Y \subseteq X$ with the image of an $n$-simplex $\sigma : \Delta ^ n \rightarrow X$, where $n > 0$. Suppose that the inverse image $\sigma ^{-1}(Y) \subseteq \Delta ^ n$ is equal to the horn $\Lambda ^{n}_{i}$ for some $0 \leq i \leq n$. Then pullback diagram of simplicial sets

3.3
\begin{equation} \begin{gathered}\label{equation:build-using-horn} \xymatrix { \Lambda ^{n}_{i} \ar [r] \ar [d] & \Delta ^ n \ar [d]^{\sigma } \\ Y \ar [r] & X } \end{gathered} \end{equation}

is also a pushout square.

**Proof.**
Fix an integer $m \geq 0$. We wish to show that $\sigma $ induces a bijection from the set of $m$-simplices of $\Delta ^{n}$ which are not contained in $\Lambda ^{n}_{i}$ to the set of $m$-simplices of $X$ which are not contained in $Y$. Surjectivity follows from our assumption that $X$ is the union of $Y$ with the image of $\sigma $. To prove injectivity, we proceed by induction on $m$. Let $\alpha ,\beta : \Delta ^{m} \rightarrow \Delta ^{n}$ be morphisms which do not factor through $\Lambda ^{n}_{i}$, and suppose that $\sigma \circ \alpha = \tau = \sigma \circ \beta $ for some simplex $\tau : \Delta ^ m \rightarrow X$; we wish to show that $\alpha = \beta $.

Suppose first that the simplex $\alpha $ is degenerate: that is, we have $\alpha (j) = \alpha (k)$ for some $0 \leq j < k \leq m$. Then $d^{m}_ j(\alpha )$ is an $(m-1)$-simplex of $\Delta ^ n$ which is not contained $\Lambda ^{n}_{i}$. It follows that $\sigma \circ d^{m}_ j(\alpha ) = d^{m}_ j(\tau ) = \sigma \circ d^{m}_ j(\beta )$ is an $(m-1)$-simplex of $X$ which is not contained in $Y$, so that $d^{m}_ j(\beta )$ is not contained in $\Lambda ^{n}_{i}$. Applying our inductive hypothesis, we deduce that $d^{m}_ j(\alpha ) = d^{m}_ j(\beta )$. The same argument shows that $d^{m}_ k(\alpha ) = d^{m}_ k(\beta )$, so that $\alpha = \beta $.

We may therefore assume without loss of generality that the simplex $\alpha $ is nondegenerate. By a similar argument, we may assume that $\beta $ is nondegenerate. The equality $\alpha = \beta $ now follows from the observation that $\Delta ^ n$ contains at most one nondegenerate $m$-simplex which is not contained in $\Lambda ^{n}_{i}$.
$\square$

Lemma 3.1.2.11. Let $n$ be a nonnegative integer. Then there exists a chain of simplicial subsets

\[ X(0) \subset X(1) \subset \cdots \subset X(n) \subset X(n+1) = \Delta ^1 \times \Delta ^ n \]

with the following properties:

- $(a)$
The simplicial $X(0)$ is given by the union of $\Delta ^1 \times \operatorname{\partial \Delta }^ n$ with $\{ 1\} \times \Delta ^ n$ (and can therefore be described abstractly as the pushout $(\Delta ^1 \times \operatorname{\partial \Delta }^ n) \coprod _{ \{ 1\} \times \operatorname{\partial \Delta }^ n} ( \{ 1 \} \times \Delta ^ n )$).

- $(b)$
For $0 \leq i \leq n$, the inclusion map $X(i) \hookrightarrow X(i+1)$ fits into a pushout diagram

\[ \xymatrix@R =50pt@C=50pt{ \Lambda ^{n+1}_{i+1} \ar [r] \ar [d] & X(i) \ar [d] \\ \Delta ^{n+1} \ar [r] & X(i+1). } \]

**Proof.**
For $0 \leq i \leq n$, let $\sigma _ i: \Delta ^{n+1} \rightarrow \Delta ^1 \times \Delta ^ n$ denote the map of simplicial sets given on vertices by the formula $\sigma _ i(j) = \begin{cases} (0,j) & \text{ if } j \leq i \\ (1,j-1) & \text{ if } j > i. \end{cases}$ We define simplicial subsets $X(i) \subseteq \Delta ^1 \times \Delta ^ n$ inductively by the formulae

\[ X(0) = (\Delta ^1 \times \operatorname{\partial \Delta }^ n) \cup (\{ 1\} \times \Delta ^ n) \quad \quad X(i+1) = X(i) \cup \operatorname{im}( \sigma _{i} ), \]

where $\operatorname{im}( \sigma _{i} )$ denotes the image of the morphism $\sigma _{i}$. Note that $\Delta ^1 \times \Delta ^ n$ is the union of the simplicial subsets $\{ \operatorname{im}(\sigma _ i) \} _{0 \leq i \leq n}$, and is therefore equal to $X(n+1)$. This definition satisfies condition $(a)$ by construction. To verify $(b)$, it will suffice to show that for $0 \leq i \leq n$, the inverse image $A_{} = \sigma _{i}^{-1} X(i)$ is equal to $\Lambda ^{n+1}_{i+1}$ (Lemma 3.1.2.10). Regarding $\sigma _{i}$ as an $(n+1)$-simplex of $\Delta ^1 \times \Delta ^ n$, we are reduced to showing that the faces $d^{n+1}_{j}( \sigma _{i} )$ belong to $X(i)$ if and only if $j \neq i+1$. One direction is clear: the face $d^{n+1}_{j}( \sigma _{i} )$ is contained in $\Delta ^1 \times \operatorname{\partial \Delta }^ n$ for $j \notin \{ i, i+1 \} $, the face $d^{n+1}_{i}(\sigma _ i) = d^{n+1}_{i}( \sigma _{i-1} )$ is contained in $\operatorname{im}( \sigma _{i-1} ) \subseteq X(i)$ for $i > 0$, and $d^{n+1}_0( \sigma _0)$ is contained in $\{ 1\} \times \Delta ^ n$. To complete the proof, it suffices to show that the face $d^{n+1}_{i+1}( \sigma _ i )$ is *not* contained in $X(i)$, which follows by inspection.
$\square$

**Proof of Proposition 3.1.2.8.**
Let us first regard the monomorphism $f': A'_{} \hookrightarrow B'_{}$ as fixed, and let $T$ be the collection of all maps $f: A_{} \rightarrow B_{}$ for which the induced map

\[ (A_{} \times B'_{} ) \coprod _{ A_{} \times A'_{} } ( B_{} \times A'_{} ) \hookrightarrow B_{} \times B'_{} \]

is anodyne. We wish to show that every anodyne morphism belongs to $T$. Since $T$ is weakly saturated, it will suffice to show that every horn inclusion $f: \Lambda ^{n}_{i} \hookrightarrow \Delta ^ n$ belongs to $T$ (for $n > 0$). Without loss of generality, we may assume that $0 < i$, so that $f$ is a retract of the map $g: (\Delta ^1 \times \Lambda ^ n_ i) \coprod _{ \{ 1\} \times \Lambda ^{n}_ i} ( \{ 1\} \times \Delta ^ n) \hookrightarrow \Delta ^1 \times \Delta ^ n$ (Lemma 3.1.2.9). It will therefore suffice to show that $g$ belongs to $T$. Replacing $f'$ by the monomorphism

\[ (\Lambda ^{n}_{i} \times B'_{} ) \coprod _{ \Lambda ^{n}_{i} \times A'_{} } (\Delta ^ n \times A'_{} ) \hookrightarrow \Delta ^{n} \times A'_{}, \]

we are reduced to showing that the inclusion $\{ 1\} \hookrightarrow \Delta ^1$ belongs to $T$.

Let $T'$ denote the collection of all morphisms of simplicial sets $f'': A''_{} \rightarrow B''_{}$ for which the map $(\{ 1\} \times B''_{} ) \coprod _{ \{ 1\} \times A''_{} } ( \Delta ^1 \times A''_{} ) \rightarrow \Delta ^1 \times B''_{}$ is anodyne. We will complete the proof by showing that $T'$ contains all monomorphisms of simplicial sets. By virtue of Proposition 1.4.5.13, it will suffice to show that $T''$ contains the inclusion map $\operatorname{\partial \Delta }^ m \hookrightarrow \Delta ^{m}$, for each $m > 0$. In other words, we are reduced to showing that the inclusion $(\{ 1\} \times \Delta ^ m ) \coprod _{ \{ 1\} \times \operatorname{\partial \Delta }^ m } ( \Delta ^1 \times \operatorname{\partial \Delta }^ m) \hookrightarrow \Delta ^1 \times \Delta ^ m$ is anodyne, which follows from Lemma 3.1.2.11.
$\square$