### 5.1.4 Covariant and Contravariant Equivalences

Fix a simplicial set $S$. In ยง5.1.3, we showed that the notion of homotopy equivalence relative to $S$ (Definition 5.1.3.1) is well-behaved when restricted to simplicial sets $X$ which are equipped with a left (or right) fibration $q: X \rightarrow S$. When working with more general objects of $(\operatorname{Set_{\Delta }})_{/S}$, it is convenient to consider a more liberal notion of equivalence.

Definition 5.1.4.1. Suppose we are given a commutative diagram of simplicial sets

\[ \xymatrix@R =50pt@C=50pt{ X \ar [rr]^{f} \ar [dr] & & Y \ar [dl] \\ & S. & } \]

We say that $f$ is a *covariant equivalence relative to $S$* if, for every left fibration $q: Z \rightarrow S$, precomposition with $f$ induces a bijection $\pi _0( \operatorname{Fun}_{/S}(Y,Z) ) \rightarrow \pi _0( \operatorname{Fun}_{/S}(X,Z) )$. We say that $f$ is a *contravariant equivalence relative to $S$* if, for every right fibration $q: Z \rightarrow S$, precomposition with $f$ induces a bijection $\pi _0(\operatorname{Fun}_{/S}(Y,Z)) \rightarrow \pi _0(\operatorname{Fun}_{/S}(X,Z))$.

Example 5.1.4.3. Suppose we are given a commutative diagram of simplicial sets

\[ \xymatrix@R =50pt@C=50pt{ X \ar [rr]^{f} \ar [dr] & & Y \ar [dl] \\ & S. & } \]

If $f$ is a homotopy equivalence relative to $S$ (Definition 5.1.3.1), then it is both a covariant and contravariant equivalence relative to $S$ (see Remark 5.1.3.2).

Example 5.1.4.4. Let $f: X \rightarrow Y$ be a morphism of simplicial sets. It follows from Proposition 4.4.2.1 that the following conditions are equivalent:

The morphism $f$ is a covariant equivalence relative to $\Delta ^{0}$.

The morphism $f$ is a contravariant equivalence relative to $\Delta ^{0}$.

The morphism $f$ is a weak homotopy equivalence.

See Proposition 5.1.4.16 for a more general statement.

Warning 5.1.4.5. The notions of covariant and contravariant equivalence are generally distinct from one another. For example, the inclusion map $i_1: \{ 1\} \hookrightarrow \Delta ^1$ is a contravariant equivalence with respect to $\Delta ^1$ (see Proposition 5.1.4.13 below). However, $i_1$ is not a covariant equivalence with respect to $\Delta ^1$: note that $i_1$ is a left fibration, but the restriction map

\[ \pi _0(\operatorname{Fun}_{/\Delta ^1}( \Delta ^1, \{ 1\} ) ) \rightarrow \pi _0( \operatorname{Fun}_{/\Delta ^1}( \{ 1\} , \{ 1\} ) ) \]

is not bijective (the left hand side is empty and the right hand side is not).

Warning 5.1.4.6. In the situation of Definition 5.1.4.1, we will often abuse terminology by referring to the morphism $f: X \rightarrow Y$ in the slice category $(\operatorname{Set_{\Delta }})_{/S}$ as a *covariant equivalence* (*contravariant equivalence*) if it is a covariant equivalence relative to $S$. Beware that there is some danger of confusion, because the notion of covariant equivalence depends on $S$. For example, the inclusion map $\{ 1\} \hookrightarrow \Delta ^1$ is not a covariant equivalence relative to $\Delta ^1$ (Warning 5.1.4.5), but *is* a covariant equivalence relative to the final object $\Delta ^{0} \in \operatorname{Set_{\Delta }}$ (Example 5.1.4.4).

Proposition 5.1.4.11. Suppose we are given a commutative diagram of simplicial sets

\[ \xymatrix@R =50pt@C=50pt{ X \ar [rr]^{f} \ar [dr] & & Y \ar [dl] \\ & S. & } \]

Let $q: Z \rightarrow S$ be a morphism of simplicial sets and let $\theta : \operatorname{Fun}_{/S}(Y,Z) \rightarrow \operatorname{Fun}_{/S}(X,Z)$ be the morphism given by precomposition with $f$. If $f$ is a covariant equivalence relative to $S$ and $q$ is a left fibration, then $\theta $ is homotopy equivalence. Similarly, if $f$ is a contravariant equivalence relative to $S$ and $q$ is a right fibration, then $\theta $ is a homotopy equivalence.

**Proof.**
We will prove the first assertion; the second follows from the same argument. We will prove that $\theta $ is a homotopy equivalence by showing that, for every simplicial set $K$, the induced map $\theta _{K}: \pi _0( \operatorname{Fun}(K, \operatorname{Fun}_{/S}(Y,Z) ) \rightarrow \pi _0( \operatorname{Fun}(K, \operatorname{Fun}_{/S}(X,Z) )$ is a bijection. Form a pullback diagram of simplicial sets

\[ \xymatrix@R =50pt@C=50pt{ Z' \ar [r] \ar [d]^{q'} & \operatorname{Fun}(K,Z) \ar [d]^{ \circ q} \\ S \ar [r] & \operatorname{Fun}(K,S). } \]

The assumption that $q$ is a left fibration guarantees that the right vertical map is also a left fibration (Corollary 4.2.3.2), so that $q': Z' \rightarrow S$ is also a left fibration. Unwinding the definitions, we see that $\theta _{K}$ can be identified with the map $\pi _0( \operatorname{Fun}_{/S}(Y, Z') ) \xrightarrow {\circ f} \pi _0( \operatorname{Fun}_{/S}(X, Z') )$, and is therefore bijective by virtue of our assumption that $f$ is a covariant equivalence relative to $S$.
$\square$

Corollary 5.1.4.12. Suppose we are given a commutative diagram of simplicial sets

\[ \xymatrix@R =50pt@C=50pt{ X \ar [rr]^{f} \ar [dr] & & Y \ar [dl] \\ & S, & } \]

where $f$ is a monomorphism. If $q: Z \rightarrow S$ is a left fibration of simplicial sets and $f$ is a covariant equivalence relative to $S$, then the induced map $\theta : \operatorname{Fun}_{/S}(Y,Z) \rightarrow \operatorname{Fun}_{/S}(X,Z)$ is a trivial Kan fibration. Similarly, if $q: Z \rightarrow S$ is a right fibration and $f$ is a contravariant equivalence relative to $S$, then $\theta $ is a trivial Kan fibration.

**Proof.**
We will prove the first assertion; the proof of the second is similar. If $q$ is a left fibration, then the morphism $\theta : \operatorname{Fun}_{/S}(Y,Z) \rightarrow \operatorname{Fun}_{/S}(X,Z)$ is a Kan fibration (Proposition 5.1.3.5). If $f$ is a covariant equivalence relative to $S$, then $\theta $ is also a homotopy equivalence (Proposition 5.1.4.11), and therefore a trivial Kan fibration (Proposition 3.3.7.4).
$\square$

Proposition 5.1.4.13. Let $f: X \hookrightarrow Y$ be a monomorphism of simplicial sets. The following conditions are equivalent:

- $(1)$
The morphism $f$ is left anodyne.

- $(2)$
The morphism $f$ is a covariant equivalence relative to $Y$.

- $(3)$
For every morphism of simplicial sets $Y \rightarrow S$, the morphism $f$ is a covariant equivalence relative to $S$.

**Proof.**
The implication $(1) \Rightarrow (2)$ follows from Remark 5.1.3.6, and the implication $(2) \Rightarrow (3)$ from Remark 5.1.4.7. We will show that $(3)$ implies $(1)$. For this, we must show that every lifting problem

\[ \xymatrix@R =50pt@C=50pt{ X \ar [d]^{f} \ar [r] & Z \ar [d]^{q} \\ Y \ar [r] \ar@ {-->}[ur] & S } \]

admits a solution, provided that $q$ is a left fibration of simplicial sets (Corollary 4.2.2.9). To prove this, it will suffice to show that the natural map $\theta : \operatorname{Fun}_{/S}(Y,Z) \rightarrow \operatorname{Fun}_{/S}(X,Z)$ is surjective on vertices. In fact, our assumption that $f$ is a monomorphism and a covariant equivalence relative to $S$ guarantees that $\theta $ is a trivial Kan fibration of simplicial sets (Corollary 5.1.4.12).
$\square$

Proposition 5.1.4.14. Suppose we are given a commutative diagram of simplicial sets

\[ \xymatrix@R =50pt@C=50pt{ X \ar [rr]^{f} \ar [dr]_{q_ X} & & Y \ar [dl]^{q_{Y}} \\ & S, & } \]

where $q_{X}$ and $q_{Y}$ are left fibrations. Then the following conditions are equivalent:

- $(1)$
The morphism $f$ is a homotopy equivalence relative to $S$.

- $(2)$
The morphism $f$ is a covariant equivalence relative to $S$.

- $(3)$
For each vertex $s \in S$, the induced map $f_{s}: X_{s} \rightarrow Y_{s}$ is a homotopy equivalence of Kan complexes.

**Proof.**
The implications $(1) \Rightarrow (2)$ and $(1) \Rightarrow (3)$ follow from Example 5.1.4.3 and Remark 5.1.3.2 (and do not require the assumption that $q_{X}$ and $q_{Y}$ are left fibrations). The implication $(3) \Rightarrow (1)$ follows from Theorem 5.1.3.3. We will complete the proof by showing that $(2) \Rightarrow (1)$. Assume that $f$ is a covariant equivalence relative to $S$. Then precomposition with $f$ induces a bijection $\pi _0( \operatorname{Fun}_{/S}(Y,X) ) \rightarrow \pi _0( \operatorname{Fun}_{/S}(X,X) )$. We can therefore choose a morphism $g: Y \rightarrow X$ in the slice category $(\operatorname{Set_{\Delta }})_{/S}$ such that $g \circ f$ and $\operatorname{id}_{X}$ belong to the same connected component of $\operatorname{Fun}_{/S}(X,X)$. Then $f \circ g \circ f$ and $f$ belong to the same connected component of the simplicial set $\operatorname{Fun}_{/S}(X,Y)$. Since the map $\pi _0( \operatorname{Fun}_{/S}(Y,Y) ) \rightarrow \pi _0( \operatorname{Fun}_{/S}(X,Y) )$ is injective, it follows that $f \circ g$ and $\operatorname{id}_{Y}$ belong to the same connected component of $\operatorname{Fun}_{/S}(Y,Y)$. It follows that $f$ is a homotopy equivalence relative to $S$ (with homotopy inverse $g$).
$\square$

We conclude by establishing a partial converse to Remark 5.1.4.8:

Proposition 5.1.4.16. Suppose we given a commutative diagram of simplicial sets

\[ \xymatrix@R =50pt@C=50pt{ X \ar [rr]^{f} \ar [dr]_{q_ X} & & Y \ar [dl]^{q_ Y} \\ & S, & } \]

where $S$ is a Kan complex. The following conditions are equivalent:

- $(1)$
The morphism $f$ is a covariant equivalence relative to $S$.

- $(2)$
The morphism $f$ is a contravariant equivalence relative to $S$.

- $(3)$
The morphism $f$ is a weak homotopy equivalence.

**Proof.**
The implications $(1) \Rightarrow (3)$ and $(2) \Rightarrow (3)$ follow from Remark 5.1.4.8 (and do not require the assumption that $S$ is a Kan complex). We will complete the proof by establishing the implication $(3) \Rightarrow (1)$ (the proof of the implication $(3) \Rightarrow (2)$ is similar). Assume that $f$ is a weak homotopy equivalence and let $Z \rightarrow S$ be a left fibration; we wish to show that the induced map $\theta : \operatorname{Fun}_{/S}(Y,Z) \rightarrow \operatorname{Fun}_{/S}(X,Z)$ is bijective on connected components. Our assumption that $S$ is a Kan complex guarantees that $q$ is a Kan fibration (Lemma 5.1.2.4), so that $Z$ is also a Kan complex. Consider the diagram of simplicial sets

5.5
\begin{equation} \begin{gathered}\label{equation:covariant-equivalence-over-Kan} \xymatrix@C =50pt@R=50pt{ \operatorname{Fun}(Y,Z) \ar [r]^-{\circ f} \ar [d]^{q \circ } & \operatorname{Fun}(X,Z) \ar [d]^{q \circ } \\ \operatorname{Fun}(Y,S) \ar [r]^-{\circ f} & \operatorname{Fun}(X,S). } \end{gathered} \end{equation}

Our assumption that $f$ is a weak homotopy equivalence guarantees that the horizontal maps in this diagram are homotopy equivalences (Corollary 3.1.6.5), so that diagram (5.5) is a homotopy Cartesian (Corollary 3.4.1.3). Corollary 3.1.3.2 guarantees that the vertical maps in the diagram (5.5) are Kan fibrations. Invoking Example 3.4.1.4, we conclude that the induced map

\[ \theta : \operatorname{Fun}_{/S}(Y,Z) = \{ q_ Y \} \times _{ \operatorname{Fun}(Y,S)} \operatorname{Fun}(Y,Z) \rightarrow \{ q_ X \} \times _{ \operatorname{Fun}(X,S) } \operatorname{Fun}(X,Z) = \operatorname{Fun}_{/S}(X,Z) \]

is also a homotopy equivalence, and is therefore bijective on connected components.
$\square$