# Kerodon

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### 1.4.4 Digression: Lifting Properties

We now review some categorical terminology which will be useful in the proof of Theorem 1.4.3.7, and in several other parts of this book.

Definition 1.4.4.1. Let $\operatorname{\mathcal{C}}$ be a category. A lifting problem in $\operatorname{\mathcal{C}}$ is a commutative diagram $\sigma :$

$\xymatrix@C =40pt@R=40pt{ A \ar [d]^{f} \ar [r]^{u} & X \ar [d]^{p} \\ B \ar [r]^{v} & Y }$

in $\operatorname{\mathcal{C}}$. A solution to the lifting problem $\sigma$ is a morphism $h: B \rightarrow X$ in $\operatorname{\mathcal{C}}$ satisfying $p \circ h = v$ and $h \circ f = u$, as indicated in the diagram

$\xymatrix@C =40pt@R=40pt{ A \ar [d]^{f} \ar [r]^{u} & X \ar [d]^{p} \\ B \ar [r]^{v} \ar [ur]^{h} & Y. }$

Remark 1.4.4.2. In the situation of Definition 1.4.4.1, we will often indicate a lifting problem by a commutative diagram

$\xymatrix@C =40pt@R=40pt{ A \ar [d]^{f} \ar [r]^{u} & X \ar [d]^{p} \\ B \ar [r]^{v} \ar@ {-->}[ur]^{h} & Y, }$

which includes a dotted arrow representing a hypothetical solution.

Definition 1.4.4.3. Let $\operatorname{\mathcal{C}}$ be a category, and suppose we are given a morphism $f: A \rightarrow B$ and $p: X \rightarrow Y$ in $\operatorname{\mathcal{C}}$. We will say that $f$ has the left lifting property with respect to $p$, or that $p$ has the right lifting property with respect to $f$, if, for every pair of morphisms $u: A \rightarrow X$ and $v: B \rightarrow Y$ satisfying $p \circ u= v \circ f$, the associated lifting problem

$\xymatrix@C =40pt@R=40pt{ A \ar [d]^{f} \ar [r]^{u} & X \ar [d]^{p} \\ B \ar [r]^{v} \ar@ {-->}[ur] & Y }$

admits a solution (that is, there exists a map $h: B \rightarrow X$ satisfying $p \circ h = v$ and $h \circ f = u$).

If $S$ is a collection of morphisms in $\operatorname{\mathcal{C}}$, we will say that a morphism $f: A \rightarrow B$ has the left lifting property with respect to $S$ if it has the left lifting property with respect to every morphism in $S$. Similarly, we will say that a morphism $p: X \rightarrow Y$ has the right lifting property with respect to $S$ if it has the right lifting property with respect to every morphism in $S$.

Let $S$ be a collection of morphisms in a category $\operatorname{\mathcal{C}}$. We now summarize some closure properties enjoyed by the collection of morphisms which have the left lifting property with respect to $S$.

Definition 1.4.4.4. Let $\operatorname{\mathcal{C}}$ be a category which admits pushouts and let $T$ be a collection of morphisms of $\operatorname{\mathcal{C}}$. We will say that $T$ is closed under pushouts if, for every pushout diagram

$\xymatrix@C =40pt@R=40pt{ A \ar [d]^{f} \ar [r] & A' \ar [d]^{f'} \\ B \ar [r] & B' }$

in the category $\operatorname{\mathcal{C}}$ where the morphism $f$ belongs to $T$, the morphism $f'$ also belongs to $T$.

Proposition 1.4.4.5. Let $\operatorname{\mathcal{C}}$ be a category which admits pushouts, let $S$ be a collection of morphisms of $\operatorname{\mathcal{C}}$, and let $T$ be the collection of all morphisms of $\operatorname{\mathcal{C}}$ having the left lifting property with respect to $S$. Then $T$ is closed under pushouts.

Proof. Suppose we are given a pushout diagram $\sigma :$

$\xymatrix@C =40pt@R=40pt{ A \ar [d]^{f} \ar [r]^{g} & A' \ar [d]^{f'} \\ B \ar [r]^{h} & B' }$

where $f$ belongs to $T$. We wish to show that $f'$ also belongs to $T$. For this, we must show that every lifting problem

$\xymatrix@C =40pt@R=40pt{ A' \ar [d]^{f'} \ar [r]^{u} & X \ar [d]^{p} \\ B' \ar [r]^{v} \ar@ {-->}[ur] & Y }$

admits a solution, provided that the morphism $p$ belongs to $S$. Using our assumption that $\sigma$ is a pushout square, we are reduced to solving the associated lifting problem

$\xymatrix@C =40pt@R=40pt{ A \ar [d]^{f} \ar [r]^{u \circ g} & X \ar [d]^{p} \\ B \ar [r]^{v \circ h} \ar@ {-->}[ur] & Y, }$

which is possible by virtue of our assumption that $f$ has the left lifting property with respect to $p$. $\square$

Definition 1.4.4.6. Let $\operatorname{\mathcal{C}}$ be a category containing a pair of objects $C$ and $C'$. We will say that $C$ is a retract of $C'$ if there exist maps $i: C \rightarrow C'$ and $r: C' \rightarrow C$ such that $r \circ i = \operatorname{id}_{C}$.

Variant 1.4.4.7. Let $\operatorname{\mathcal{C}}$ be a category. We will say that a morphism $f: C \rightarrow D$ of $\operatorname{\mathcal{C}}$ is a retract of another morphism $f': C' \rightarrow D'$ if it is a retract of $f'$ when viewed as an object of the functor category $\operatorname{Fun}( , \operatorname{\mathcal{C}})$. In other words, we say that $f$ is a retract of $f'$ if there exists a commutative diagram

$\xymatrix@C =40pt@R=40pt{ C \ar [r]^{i} \ar [d]^{f} & C' \ar [d]^{f'} \ar [r]^{r} & C \ar [d]^{f} \\ D \ar [r]^{\overline{i}} & D' \ar [r]^{\overline{r}} & D }$

in the category $\operatorname{\mathcal{C}}$, where $r \circ i = \operatorname{id}_{C}$ and $\overline{r} \circ \overline{i} = \operatorname{id}_{D}$.

We say that a collection of morphisms $T$ of $\operatorname{\mathcal{C}}$ is closed under retracts if, for every pair of morphisms $f,f'$ in $\operatorname{\mathcal{C}}$, if $f$ is a retract of $f'$ and $f'$ belongs to $T$, then $f$ also belongs to $T$.

Exercise 1.4.4.8. Let $\operatorname{\mathcal{C}}$ be a category and let $T$ be the collection of all monomorphisms in $\operatorname{\mathcal{C}}$. Show that $T$ is closed under retracts.

Proposition 1.4.4.9. Let $\operatorname{\mathcal{C}}$ be a category, let $S$ be a collection of morphisms of $\operatorname{\mathcal{C}}$, and let $T$ be the collection of all morphisms of $\operatorname{\mathcal{C}}$ having the left lifting property with respect to $S$. Then $T$ is closed under retracts.

Proof. Let $f'$ be a morphism of $\operatorname{\mathcal{C}}$ which belongs to $T$ and let $f$ be a retract of $f'$, so that there exists a commutative diagram

$\xymatrix@C =40pt@R=40pt{ C \ar [r]^{i} \ar [d]^{f} & C' \ar [d]^{f'} \ar [r]^{r} & C \ar [d]^{f} \\ D \ar [r]^{\overline{i}} & D' \ar [r]^{\overline{r}} & D }$

with $r \circ i = \operatorname{id}_{C}$ and $\overline{r} \circ \overline{i} = \operatorname{id}_{D}$. We wish to show that $f$ also belongs to $T$. Consider a lifting problem $\sigma :$

$\xymatrix@C =40pt@R=40pt{ C \ar [d]^{f} \ar [r]^{u} & X \ar [d]^{p} \\ D \ar [r]^{v} \ar@ {-->}[ur]^{h} & Y, }$

where $p$ belongs to $S$. Our assumption $f' \in T$ ensures that the associated lifting problem

$\xymatrix@C =40pt@R=40pt{ C' \ar [d]^{f'} \ar [r]^{u \circ r} & X \ar [d]^{p} \\ D' \ar [r]^{v \circ \overline{r}} \ar@ {-->}[ur] & Y }$

admits a solution: that is, we can choose a morphism $h': D' \rightarrow X$ satisfying $p \circ h' = v \circ \overline{r}$ and $h' \circ f' = u \circ r$. Then the morphism $h = h' \circ \overline{i}$ is a solution to the lifting problem $\sigma$, by virtue of the calculations

$p \circ h = p \circ h' \circ \overline{i} = v \circ \overline{r} \circ \overline{i} = v$

$h \circ f = h' \circ \overline{i} \circ f = h' \circ f' \circ i = u \circ r \circ i = u.$
$\square$

Definition 1.4.4.10. For every ordinal number $\alpha$, let $[ \alpha ] = \{ \beta : \beta \leq \alpha \}$ denote the collection of all ordinal numbers which are less than or equal to $\alpha$, regarded as a linearly ordered set.

Let $\operatorname{\mathcal{C}}$ be a category and let $T$ be a collection of morphisms of $\operatorname{\mathcal{C}}$. We will say that a morphism $f$ of $\operatorname{\mathcal{C}}$ is a transfinite composition of morphisms of $T$ if there exists an ordinal number $\alpha$ and a functor $F: [ \alpha ] \rightarrow \operatorname{\mathcal{C}}$, given by a collection of objects $\{ C_{\beta } \} _{ \beta \leq \alpha }$ and morphisms $\{ f_{\gamma , \beta }: C_{\beta } \rightarrow C_{\gamma } \} _{\beta \leq \gamma }$ with the following properties:

$(a)$

For every nonzero limit ordinal $\lambda \leq \alpha$, the functor $F$ exhibits $C_{\lambda }$ as a colimit of the diagram $( \{ C_{\beta } \} _{\beta < \lambda }, \{ f_{\gamma , \beta } \} _{\beta \leq \gamma < \lambda } )$.

$(b)$

For every ordinal $\beta < \alpha$, the morphism $f_{\beta +1, \beta }$ belongs to $T$.

$(c)$

The morphism $f$ is equal to $f_{\alpha ,0}: C_0 \rightarrow C_{\alpha }$.

We will say that $T$ is closed under transfinite composition if, for every morphism $f$ which is a transfinite composition of morphisms of $T$, we have $f \in T$.

Example 1.4.4.11. Let $\operatorname{\mathcal{C}}$ be a category and let $T$ be a collection of morphisms of $\operatorname{\mathcal{C}}$. Then every identity morphism of $\operatorname{\mathcal{C}}$ is a transfinite composition of morphisms of $T$ (take $\alpha = 0$ in Definition 1.4.4.10). In particular, if $T$ is closed under transfinite composition, then it contains every identity morphism of $\operatorname{\mathcal{C}}$.

Example 1.4.4.12. Let $\operatorname{\mathcal{C}}$ be a category and let $T$ be a collection of morphisms of $\operatorname{\mathcal{C}}$. Then every morphism of $T$ is a transfinite composition of morphisms of $T$ (take $\alpha = 1$ in Definition 1.4.4.10).

Example 1.4.4.13. Let $\operatorname{\mathcal{C}}$ be a category and let $T$ be a collection of morphisms of $\operatorname{\mathcal{C}}$ which contains a pair of composable morphisms $f: C_0 \rightarrow C_1$ and $g: C_1 \rightarrow C_2$. Then the composition $g \circ f$ is a transfinite composition of morphisms of $\operatorname{\mathcal{C}}$ (take $\alpha = 2$ in Definition 1.4.4.10). In particular, if $T$ is closed under transfinite composition, then it is closed under composition.

Proposition 1.4.4.14. Let $\operatorname{\mathcal{C}}$ be a category, let $S$ be a collection of morphisms in $\operatorname{\mathcal{C}}$, and let $T$ be the collection of all morphisms of $\operatorname{\mathcal{C}}$ which have the left lifting property with respect to $S$. Then $T$ is closed under transfinite composition.

Proof. Let $\alpha$ be an ordinal and suppose we are given a functor $[\alpha ] \rightarrow \operatorname{\mathcal{C}}$, given by a pair

$( \{ C_{\beta } \} _{\beta \leq \alpha }, \{ f_{\gamma ,\beta } \} _{\beta \leq \gamma \leq \alpha } )$

which satisfies condition $(a)$ of Definition 1.4.4.10. Assume that each of the morphisms $f_{\beta +1,\beta }$ belongs to $T$. We wish to show that the morphism $f_{\alpha ,0}$ also belongs to $T$. For this, we must show that every lifting problem $\sigma :$

$\xymatrix@C =40pt@R=40pt{ C_0 \ar [d]_{ f_{\alpha ,0} } \ar [r]^{u} & X \ar [d]^{p} \\ C_{\alpha } \ar@ {-->}[ur] \ar [r]^{v} & Y }$

admits a solution, provided that $p$ belongs to $S$. We construct a collection of morphisms $\{ u_{\beta }: C_{\beta } \rightarrow X \} _{\beta \leq \alpha }$, satisfying the requirements $p \circ u_{\beta } = v \circ f_{\alpha ,\beta }$ and $u_{\beta } = u_{\gamma } \circ f_{\gamma ,\beta }$ for $\beta \leq \gamma$, using transfinite recursion. Fix an ordinal $\gamma \leq \alpha$, and assume that the morphisms $\{ u_{\beta } \} _{ \beta < \gamma }$ have been constructed. We consider three cases:

• If $\gamma = 0$, we set $u_{\gamma } = u$.

• If $\gamma$ is a nonzero limit ordinal, then our hypothesis that $C_{\gamma }$ is the colimit of the diagram $\{ C_{\beta } \} _{\beta < \gamma }$ guarantees that there is a unique morphism $u_{\gamma }: C_{\gamma } \rightarrow X$ satisfying $u_{\beta } = u_{\gamma } \circ f_{\gamma ,\beta }$ for $\beta < \gamma$. Moreover, our assumption that the equality $p \circ u_{\beta } = v \circ f_{\alpha ,\beta }$ holds for $\beta < \gamma$ guarantees that it also holds for $\beta = \gamma$.

• Suppose that $\gamma = \beta +1$ is a successor ordinal. In this case, we take $u_{\gamma }$ to be any solution to the lifting problem

$\xymatrix@C =70pt@R=70pt{ C_{\beta } \ar [d]_{ f_{\beta +1,\beta } } \ar [r]^{u_{\beta }} & X \ar [d]^{p} \\ C_{\beta +1} \ar@ {-->}[ur] \ar [r]^{v \circ f_{\alpha , \beta +1}} & Y, }$

which exists by virtue of our assumption that $f_{\beta +1,\beta }$ belongs to $T$.

We now complete the proof by observing that $u_{\alpha }$ is a solution to the lifting problem $\sigma$. $\square$

Motivated by the preceding discussion, we introduce the following:

Definition 1.4.4.15. Let $\operatorname{\mathcal{C}}$ be a category which admits small colimits and let $T$ be a collection of morphisms of $\operatorname{\mathcal{C}}$. We will say that $T$ is weakly saturated if it is closed under pushouts (Definition 1.4.4.4), retracts (Variant 1.4.4.7), and transfinite composition (Definition 1.4.4.10).

Proposition 1.4.4.16. Let $\operatorname{\mathcal{C}}$ be a category which admits small colimits, let $S$ be a collection of morphisms of $\operatorname{\mathcal{C}}$, and let $T$ be the collection of all morphisms of $\operatorname{\mathcal{C}}$ which have the left lifting property with respect to $S$. Then $T$ is weakly saturated.

Remark 1.4.4.17. Let $\operatorname{\mathcal{C}}$ be a category and let $T_0$ be a collection of morphisms of $\operatorname{\mathcal{C}}$. Then there exists a smallest collection of morphisms $T$ of $\operatorname{\mathcal{C}}$ such that $T_0 \subseteq T$ and $T$ is weakly saturated (for example, we can take $T$ to be the intersection of all the weakly saturated collections of morphisms containing $T_0$). We will refer to $T$ as the weakly saturated collection of morphisms generated by $T_0$. It follows from Proposition 1.4.4.16 that if every morphism of $T_0$ has the left lifting property with respect to some collection of morphisms $S$, then every morphism of $T$ also has the left lifting property with respect to $S$.