nLab injective or projective morphism

Idea

Injective and projective morphisms are morphisms in a category satisfying some lifting property. Often they appear jointly to form weak factorization systems.

Definition

Injective and projective

Given a category $\mathcal{C}$ and a class $J \subset Mor(\mathcal{C})$ of its morphisms, then any morphism $f \in Mor(\mathcal{C})$ is called

• a $J$-injective morphism if it has the right lifting property against elements in $J$;

• a $J$-projective morphism if it has the left lifting property against elements in $J$.

If $\mathcal{C}$ has a terminal object $\ast$, then an object $X$ for which $X \stackrel{\exists!}{\to} \ast$ is a $J$-injective morphism is called a $J$-injective object.

If $\mathcal{C}$ has an initial object $\emptyset$, then an object $X$ for which $\emptyset \stackrel{\exists!}{\to} X$ is a $J$-projective morphism is called a $J$-projective object.

Fibration and cofibration

Frequently one furthermore says:

• a $J$-cofibration is a morphism that has the left lifting property against all $J$-injective morphisms;

• a $J$-fibration is a morphism that has the right lifting property against all $J$-projective morphisms.

Properties

Closure properties

Proposition

Let $\mathcal{C}$ be a category and let $K\subset Mor(\mathcal{C})$ be a class of morphisms. Write $K Proj$ and $K Inj$, respectively, for the sub-classes of $K$-projective morphisms and of $K$-injective morphisms (the “Quillen negation” of $K$). Then:

1. Both classes contain the class of isomorphism of $\mathcal{C}$.

2. Both classes are closed under composition in $\mathcal{C}$.

   $K Proj$ is also closed under transfinite composition.

3. Both classes are closed under forming retracts in the arrow category $\mathcal{C}^{\Delta[1]}$.

4. $K Proj$ is closed under forming pushouts of morphisms in $\mathcal{C}$ (“cobase change”).

   $K Inj$ is closed under forming pullback of morphisms in $\mathcal{C}$ (“base change”).

5. $K Proj$ is closed under forming coproducts in $\mathcal{C}^{\Delta[1]}$.

   $K Inj$ is closed under forming products in $\mathcal{C}^{\Delta[1]}$.

Proof

We go through each item in turn.

containing isomorphisms

Given a commuting square

$$\array{ A &\overset{f}{\rightarrow}& X \\ {}_{\mathllap{\in Iso}}^{\mathllap{i}}\downarrow && \downarrow^{\mathrlap{p}} \\ B &\underset{g}{\longrightarrow}& Y }$$

with the left morphism an isomorphism, the a lift is given by using the inverse of this isomorphism ${}^{{f \circ i^{-1}}}\nearrow$. Hence in particular there is a lift when $p \in K$ and so $i \in K Proj$. The other case is formally dual.

closure under composition

Given a commuting square of the form

$$\array{ A &\longrightarrow& X \\ \downarrow && \downarrow^{\mathrlap{p_1}}_{\mathrlap{\in K Inj}} \\ {}^{\mathllap{i}}_{\mathllap{\in K}}\downarrow && \downarrow^{\mathrlap{p_2}}_{\mathrlap{\in K Inj}} \\ B &\longrightarrow& Y }$$

consider its pasting decomposition as

$$\array{ A &\longrightarrow& X \\ \downarrow &\searrow& \downarrow^{\mathrlap{p_1}}_{\mathrlap{\in K Inj}} \\ {}^{\mathllap{i}}_{\mathllap{\in K}}\downarrow && \downarrow^{\mathrlap{p_2}}_{\mathrlap{\in K Inj}} \\ B &\longrightarrow& Y } \,.$$

Now the bottom commuting square has a lift, by assumption. This yields another pasting decomposition

$$\array{ A &\longrightarrow& X \\ {}^{\mathllap{i}}_{\mathllap{\in K}}\downarrow && \downarrow^{\mathrlap{p_1}}_{\mathrlap{\in K Inj}} \\ \downarrow &\nearrow& \downarrow^{\mathrlap{p_2}}_{\mathrlap{\in K Inj}} \\ B &\longrightarrow& Y }$$

and now the top commuting square has a lift by assumption. This is now equivalently a lift in the total diagram, showing that $p_1\circ p_1$ has the right lifting property against $K$ and is hence in $K Inj$. The case of composing two morphisms in $K Proj$ is formally dual. From this the closure of $K Proj$ under transfinite composition follows since the latter is given by colimits of sequential composition and successive lifts against the underlying sequence as above constitutes a cocone, whence the extension of the lift to the colimit follows by its universal property.

closure under retracts

Let $j$ be the retract of an $i \in K Proj$, i.e. let there be a commuting diagram of the form.

$$\array{ id_A \colon & A &\longrightarrow& C &\longrightarrow& A \\ & \downarrow^{\mathrlap{j}} && \downarrow^{\mathrlap{i}}_{\mathrlap{\in K Proj}} && \downarrow^{\mathrlap{j}} \\ id_B \colon & B &\longrightarrow& D &\longrightarrow& B } \,.$$

Then for

$$\array{ A &\longrightarrow& X \\ {}^{\mathllap{j}}\downarrow && \downarrow^{\mathrlap{f}}_{\mathrlap{\in K}} \\ B &\longrightarrow& Y }$$

a commuting square, it is equivalent to its pasting composite with that retract diagram

$$\array{ A &\longrightarrow& C &\longrightarrow& A &\longrightarrow& X \\ \downarrow^{\mathrlap{j}} && \downarrow^{\mathrlap{i}}_{\mathrlap{\in K Proj}} && \downarrow^{\mathrlap{j}} && \downarrow^{\mathrlap{f}}_{\mathrlap{\in K}} \\ B &\longrightarrow& D &\longrightarrow& B &\longrightarrow & Y } \,.$$

Now the pasting composite of the two squares on the right has a lift, by assumption,

$$\array{ A &\longrightarrow& C &\longrightarrow& A &\longrightarrow& X \\ \downarrow^{\mathrlap{j}} && \downarrow^{\mathrlap{i}}_{\mathrlap{\in K Proj}} && \nearrow && \downarrow^{\mathrlap{f}}_{\mathrlap{\in K}} \\ B &\longrightarrow& D &\longrightarrow& B &\longrightarrow & Y } \,.$$

By composition, this is also a lift in the total outer rectangle, hence in the original square. Hence $j$ has the left lifting property against all $p \in K$ and hence is in $K Proj$. The other case is formally dual.

closure under pushout and pullback

Let $p \in K Inj$ and and let

$$\array{ Z \times_f X &\longrightarrow& X \\ {}^{\mathllap{{f^* p}}}\downarrow && \downarrow^{\mathrlap{p}} \\ Z &\stackrel{f}{\longrightarrow} & Y }$$

be a pullback diagram in $\mathcal{C}$. We need to show that $f^* p$ has the right lifting property with respect to all $i \in K$. So let

$$\array{ A &\longrightarrow& Z \times_f X \\ {}^{\mathllap{i}}_{\mathllap{\in K}}\downarrow && \downarrow^{\mathrlap{\mathrlap{f^* p}}} \\ B &\stackrel{g}{\longrightarrow}& Z }$$

be a commuting square. We need to construct a diagonal lift of that square. To that end, first consider the pasting composite with the pullback square from above to obtain the commuting diagram

$$\array{ A &\longrightarrow& Z \times_f X &\longrightarrow& X \\ {}^{\mathllap{i}}\downarrow && \downarrow^{\mathrlap{f^* p}} && \downarrow^{\mathrlap{p}} \\ B &\stackrel{g}{\longrightarrow}& Z &\stackrel{f}{\longrightarrow}& Y } \,.$$

By the right lifting property of $p$, there is a diagonal lift of the total outer diagram

$$\array{ A &\longrightarrow& X \\ \downarrow^{\mathrlap{i}} &{}^{\hat {(f g)}}\nearrow& \downarrow^{\mathrlap{p}} \\ B &\stackrel{f g}{\longrightarrow}& Y } \,.$$

By the universal property of the pullback this gives rise to the lift $\hat g$ in

$$\array{ && Z \times_f X &\longrightarrow& X \\ &{}^{\hat g} \nearrow& \downarrow^{\mathrlap{f^* p}} && \downarrow^{\mathrlap{p}} \\ B &\stackrel{g}{\longrightarrow}& Z &\stackrel{f}{\longrightarrow}& Y } \,.$$

In order for $\hat g$ to qualify as the intended lift of the total diagram, it remains to show that

$$\array{ A &\longrightarrow& Z \times_f X \\ \downarrow^{\mathrlap{i}} & {}^{\hat g}\nearrow \\ B }$$

commutes. To do so we notice that we obtain two cones with tip $A$:

• one is given by the morphisms

  1. $A \to Z \times_f X \to X$
  2. $A \stackrel{i}{\to} B \stackrel{g}{\to} Z$

  with universal morphism into the pullback being

  • $A \to Z \times_f X$

• the other by

  1. $A \stackrel{i}{\to} B \stackrel{\hat g}{\to} Z \times_f X \to X$
  2. $A \stackrel{i}{\to} B \stackrel{g}{\to} Z$.

  with universal morphism into the pullback being

  • $A \stackrel{i}{\to} B \stackrel{\hat g}{\to} Z \times_f X$.

The commutativity of the diagrams that we have established so far shows that the first and second morphisms here equal each other, respectively. By the fact that the universal morphism into a pullback diagram is unique this implies the required identity of morphisms.

The other case is formally dual.

closure under (co-)products

Let $\{(A_s \overset{i_s}{\to} B_s) \in K Proj\}_{s \in S}$ be a set of elements of $K Proj$. Since colimits in the presheaf category $\mathcal{C}^{\Delta[1]}$ are computed componentwise, their coproduct in this arrow category is the universal morphism out of the coproduct of objects $\underset{s \in S}{\coprod} A_s$ induced via its universal property by the set of morphisms $i_s$:

$$\underset{s \in S}{\sqcup} A_s \overset{(i_s)_{s\in S}}{\longrightarrow} \underset{s \in S}{\sqcup} B_s \,.$$

Now let

$$\array{ \underset{s \in S}{\sqcup} A_s &\longrightarrow& X \\ {}^{\mathllap{(i_s)_{s\in S}}}\downarrow && \downarrow^{\mathrlap{f}}_{\mathrlap{\in K}} \\ \underset{s \in S}{\sqcup} B_s &\longrightarrow& Y }$$

be a commuting square. This is in particular a cocone under the coproduct of objects, hence by the universal property of the coproduct, this is equivalent to a set of commuting diagrams

$$\left\{ \;\;\;\;\;\;\;\;\; \array{ A_s &\longrightarrow& X \\ {}^{\mathllap{i_s}}_{\mathllap{\in K Proj}}\downarrow && \downarrow^{\mathrlap{f}}_{\mathrlap{\in K}} \\ B_s &\longrightarrow& Y } \;\;\;\; \right\}_{s\in S} \,.$$

By assumption, each of these has a lift $\ell_s$. The collection of these lifts

$$\left\{ \;\;\;\;\;\;\;\;\; \array{ A_s &\longrightarrow& X \\ {}^{\mathllap{i_s}}_{\mathllap{\in Proj}}\downarrow &{}^{\ell_s}\nearrow& \downarrow^{\mathrlap{f}}_{\mathrlap{\in K}} \\ B_s &\longrightarrow& Y } \;\;\;\; \right\}_{s\in S}$$

is now itself a compatible cocone, and so once more by the universal property of the coproduct, this is equivalent to a lift $(\ell_s)_{s\in S}$ in the original square

$$\array{ \underset{s \in S}{\sqcup} A_s &\longrightarrow& X \\ {}^{\mathllap{(i_s)_{s\in S}}}\downarrow &{}^{(\ell_s)_{s\in S}}\nearrow& \downarrow^{\mathrlap{f}}_{\mathrlap{\in K}} \\ \underset{s \in S}{\sqcup} B_s &\longrightarrow& Y } \,.$$

This shows that the coproduct of the $i_s$ has the left lifting property against all $f\in K$ and is hence in $K Proj$. The other case is formally dual.

Related concepts

• the small object argument is about factoring morphisms into $J$-cofibrations followed by $J$-injective morphisms.