# Kerodon

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## 8.5 Reclassified Tags

This section consists of tags which were replaced by a tag with similar content but different environment (for example, a lemma which became a theorem).

Proposition 8.5.0.42. Let $f: X_{} \rightarrow S_{}$ be a morphism of simplicial sets. Then $f$ is a Kan fibration if and only if it is both a left fibration and a right fibration.

Variant 8.5.0.43 (Strictly Unitary $2$-Categories). The contents of this tag can be found in Definition 2.2.7.1.

We say that a $2$-category $\operatorname{\mathcal{C}}$ is strictly unitary if, for every $1$-morphism $f: X \rightarrow Y$ in $\operatorname{\mathcal{C}}$, we have equalities

$\operatorname{id}_{Y} \circ f = f = f \circ \operatorname{id}_{X},$

and the left and right unit constraints $\lambda _{f}$, $\rho _{f}$ are the identity $2$-morphisms from $f$ to itself. Every strict $2$-category is strictly unitary, but the converse is false: we will see later that every $2$-category is isomorphic (in an appropriate sense) to a strictly unitary $2$-category (see Example 8.5.0.47).

Example 8.5.0.45. The contents of this tag can now be found in Remark 2.2.7.3.

Let $\operatorname{\mathcal{C}}$ be a strictly unitary $2$-category (Variant 8.5.0.43). Then Proposition 2.2.1.16 can be formulated more simply as follows: for every pair of composable $1$-morphisms $X \xrightarrow {f} Y \xrightarrow {g} Z$, the associativity constraints $\alpha _{ \operatorname{id}_ Z, g, f}$ and $\alpha _{g,f,\operatorname{id}_ X}$ are equal to the identity (as $2$-morphisms from $g \circ f$ to itself).

Example 8.5.0.46. The contents of this tag are now contained in Proposition 1.3.5.7.

Let $\operatorname{\mathcal{C}}$ be an $\infty$-category. Then the homotopy category $\mathrm{h} \mathit{\operatorname{\mathcal{C}}}$ constructed in Definition 1.3.5.3 is also a homotopy category of $\operatorname{\mathcal{C}}$ in the sense of Definition 1.2.5.1. More precisely, the map $u: \operatorname{\mathcal{C}}\rightarrow \operatorname{N}_{\bullet }( \mathrm{h} \mathit{\operatorname{\mathcal{C}}} )$ of Construction 1.3.5.6 exhibits $\mathrm{h} \mathit{\operatorname{\mathcal{C}}}$ as a homotopy category of $\operatorname{\mathcal{C}}$, by virtue of Proposition 1.3.5.7.

Example 8.5.0.47. The contents of this tag are now contained in Remark 2.2.7.6 and Proposition 2.2.7.7.

Let $\operatorname{\mathcal{C}}$ be any $2$-category. Then the left and right unit constraints on $\operatorname{\mathcal{C}}$ determine a twisting cochain $\{ \mu _{g,f} \}$, given concretely by the formula

$\mu _{g,f} = \begin{cases} \lambda _{f}: g \circ f \Rightarrow f & \text{ if } g = \operatorname{id}_{Y} \\ \rho _{g}: g \circ f \Rightarrow g & \text{ if } f = \operatorname{id}_ Y \\ \operatorname{id}_{g \circ f}: g \circ f \Rightarrow g \circ f & \text{ otherwise. } \end{cases}$

Note that this prescription is consistent, since $\lambda _{f} = \rho _{g}$ in the special case where $f = g = \operatorname{id}_{Y}$ (Corollary 2.2.1.15).

Let $\operatorname{\mathcal{C}}'$ be the twist of $\operatorname{\mathcal{C}}$ with respect to the cocycle $\{ \mu _{g,f} \}$. Then $\operatorname{\mathcal{C}}'$ is a strictly unitary $2$-category (in the sense of Variant 8.5.0.43), and Exercise 2.2.6.9 supplies a strictly unitary isomorphism of $2$-categories $\operatorname{\mathcal{C}}\simeq \operatorname{\mathcal{C}}'$ In particular, for every $2$-category $\operatorname{\mathcal{C}}$, there exists a strictly unitary isomorphism of $2$-categories $\operatorname{\mathcal{C}}\simeq \operatorname{\mathcal{C}}'$, where $\operatorname{\mathcal{C}}'$ is strictly unitary.

Proposition 8.5.0.48. Let $i: A_{} \hookrightarrow B_{}$ be an anodyne morphism of simplicial sets and let $f: X_{} \rightarrow S_{}$ be a Kan fibration. Then the induced map

$\operatorname{Fun}( B_{}, X_{} ) \rightarrow \operatorname{Fun}( B_{}, S_{} ) \times _{ \operatorname{Fun}( A_{}, S_{} )} \operatorname{Fun}( A_{}, X_{} )$

is a trivial Kan fibration.

Lemma 8.5.0.49. Let $f: A_{} \hookrightarrow B_{}$ and $f': A'_{} \hookrightarrow B'_{}$ be monomorphisms of simplicial sets. If either $f$ is anodyne, then the induced map

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

is anodyne.

Proposition 8.5.0.50. Let $f: X_{} \rightarrow S_{}$ be a morphism of simplicial sets. The following conditions are equivalent:

$(1)$

The morphism $f$ is a right fibration.

$(2)$

For every pair of integers $0 < i \leq n$, every lifting problem

$\xymatrix@R =50pt@C=50pt{ \Lambda ^{n}_{i} \ar [r]^-{\sigma _0} \ar [d] & X_{} \ar [d]^{f} \\ \Delta ^{n} \ar@ {-->}[ur]^-{\sigma } \ar [r]^-{ \overline{\sigma } } & S_{} }$

admits a solution (indicated by the dotted arrow).

Lemma 8.5.0.51. Let $q: X \rightarrow S$ be a left fibration of simplicial sets, where $S$ is a Kan complex. Suppose that $q$ induces a surjection $\pi _0(q): \pi _0(X) \rightarrow \pi _0(S)$. Then $q$ is surjective on vertices.

Proof. Fix a vertex $s \in S$. Since $\pi _0(q)$ is surjective, there exists a vertex $x \in X$ for which $q(x)$ and $s$ belong to the same connected component of $\pi _0(S)$. Since $S$ is a Kan complex, we can choose an edge $e: q(x) \rightarrow s$ in the simplicial set $S$. Our assumption that $q$ is a left fibration guarantees that we can write $e = q(\overline{e})$ for some edge $\overline{e}: x \rightarrow \overline{s}$ of the simplicial set $X$. In particular, there exists a vertex $\overline{s} \in X$ satisfying $q( \overline{s} ) = s$. $\square$

Remark 8.5.0.52. Assuming Theorem 5.2.2.15, one can give a more direct proof of ***. Let $q: X \rightarrow S$ be a left fibration of simplicial sets, where $S$ is a Kan complex. To show that $q$ is a Kan fibration, it will suffice (by virtue of Theorem 5.2.2.15) to show that the covariant transport functor $\mathrm{h} \mathit{S} \rightarrow \mathrm{h} \mathit{\operatorname{Kan}}$ carries every morphism in $\mathrm{h} \mathit{S}$ to an invertible morphism in $\mathrm{h} \mathit{\operatorname{Kan}}$. This is clear, since the homotopy category $\mathrm{h} \mathit{S}$ is a groupoid (Proposition 1.3.6.10).

Remark 8.5.0.53. Assuming Theorem 5.2.2.15, one can give a more direct proof of Proposition 4.4.2.14. Let $q: X \rightarrow S$ be a left fibration of simplicial sets with the property that, for every vertex $s \in S$, the fiber $X_{s}$ is a contractible Kan complex. For each edge $e: s \rightarrow s'$ of $S$, the covariant transport $e_{!}: X_{s} \rightarrow X_{s'}$ is a morphism between contractible Kan complexes, and is therefore automatically a homotopy equivalence. Applying Theorem 5.2.2.15, we deduce that $q$ is a Kan fibration. Since the fibers of $q$ are contractible, Proposition 3.3.7.4 guarantees that $q$ is a trivial Kan fibration.