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10.2.3 Images

Let $X$ and $Y$ be sets. Recall that the image of a function $f: X \rightarrow Y$ is defined to be the subset $\operatorname{im}(f) = \{ y \in Y: f^{-1}\{ y\} \neq \emptyset \}$. More abstractly, the set $\operatorname{im}(f)$ is characterized (up to isomorphism) by the requirement that $f$ factors as a composition

$X \stackrel{q}{\twoheadrightarrow } \operatorname{im}(f) \stackrel{i}{\hookrightarrow } Y,$

where $q$ is surjective and $i$ is injective. This motivates the following:

Definition 10.2.3.1. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category, let $Y$ be an object of $\operatorname{\mathcal{C}}$, and let $Y_0 \subseteq Y$ be a subobject: that is, an object of $\operatorname{\mathcal{C}}$ equipped with a (specified) monomorphism $i: Y_0 \hookrightarrow Y$ (see Definition 9.2.4.25). We will say that $Y_0$ is an image of a morphism $f: X \rightarrow Y$ if the homotopy class $[f]$ factors as a composition $[i] \circ [q]$, where $q: X \twoheadrightarrow Y_0$ is a quotient morphism in $\operatorname{\mathcal{C}}$.

Remark 10.2.3.2. In the situation of Definition 10.2.3.1, our assumption that $i$ is a monomorphism guarantees that the composition map $\operatorname{Hom}_{\mathrm{h} \mathit{\operatorname{\mathcal{C}}}}( X, Y_0 ) \xrightarrow { [i] \circ } \operatorname{Hom}_{\mathrm{h} \mathit{\operatorname{\mathcal{C}}}}( X, Y )$ is injective. It follows that if there exists a morphism $q: X \rightarrow Y_0$ satisfying $[f] = [i] \circ [q]$, then $q$ is uniquely determined up to homotopy. In particular, the condition that $q$ is a quotient morphism is independent of the choice of $q$ (see Example 10.2.2.13).

Remark 10.2.3.3. In the situation of Definition 10.2.3.1, $Y_0$ is an image of $f$ if and only if there exists a $2$-simplex

10.22
$$\begin{gathered}\label{equation:image-in-infinity} \xymatrix@R =50pt@C=50pt{ & Y_0 \ar [dr]^{i} & \\ X \ar [ur]^{q} \ar [rr]^{f} & & Y } \end{gathered}$$

in the $\infty$-category $\operatorname{\mathcal{C}}$, where $q$ is a quotient morphism. If this condition is satisfied, we say that the $2$-simplex (10.22) exhibits $Y_0$ as an image of $f$.

Example 10.2.3.4 (Images of Sets). Let $f: X \rightarrow Y$ be a function between sets, and set $Y_0 = \{ y \in Y: f^{-1} \{ y\} \neq \emptyset \}$. Then $f$ determines a surjection from $X$ to $Y_0$, which is a quotient morphism in the category of sets (Example 10.2.2.8). It follows that the commutative diagram

$\xymatrix@R =50pt@C=50pt{ & Y_0 \ar [dr]^{i} & \\ X \ar [ur]^{f} \ar [rr]^{f} & & Y }$

exhibits $Y_0$ as an image of $f$.

Example 10.2.3.5 (Images of Monomorphisms). Let $\operatorname{\mathcal{C}}$ be an $\infty$-category and let $f: X \hookrightarrow Y$ be a monomorphism in $\operatorname{\mathcal{C}}$. Since the identity map $\operatorname{id}_{X}$ is a quotient morphism (Exercise 10.2.2.3), the left-degenerate $2$-simplex

$\xymatrix@R =50pt@C=50pt{ & X \ar [dr]^{f} & \\ X \ar [ur]^{\operatorname{id}_ X} \ar [rr]^{f} & & Y }$

exhibits $X$ as an image of $f$.

Proposition 10.2.3.6 (Images of Quotient Morphisms). Let $\operatorname{\mathcal{C}}$ be an $\infty$-category and let $f: X \rightarrow Y$ be a morphism of $\operatorname{\mathcal{C}}$. Then $f$ is a quotient morphism if and only if $Y$ is an image of $f$ (when regarded as a subobject of itself).

Proof. Assume first that $f$ is a quotient morphism. Since the identity map $\operatorname{id}_{Y}$ is a monomorphism (Example 9.2.4.9), the right-degenerate $2$-simplex

$\xymatrix@R =50pt@C=50pt{ & Y \ar [dr]^{\operatorname{id}_ Y} & \\ X \ar [ur]^{f} \ar [rr]^{f} & & Y }$

exhibits $Y$ as an image of $f$. Conversely, if $Y$ is an image of $f$, then $f$ factors as the composition of a quotient morphism and an isomorphism, and is therefore a quotient morphism by virtue of Corollary 10.2.2.12. $\square$

Warning 10.2.3.7. The terminology of Definition 10.2.3.1 is not entirely standard. In the setting of additive categories, many authors refer to an object $Y$ as the image of a morphism $f: X \rightarrow Y$ if it is a kernel of the tautological map $Y \twoheadrightarrow \operatorname{coker}(f)$. This agrees with Definition 10.2.3.1 when $\operatorname{\mathcal{C}}$ is an abelian category (Proposition ), but not in general.

Warning 10.2.3.8 (Essential Images). Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be a functor of $\infty$-categories. Recall that the essential image of $F$ is the full subcategory $\operatorname{\mathcal{D}}_0 \subseteq \operatorname{\mathcal{D}}$ spanned by objects $D \in \operatorname{\mathcal{D}}$ which are isomorphic to $F(C)$, for some object $C \in \operatorname{\mathcal{C}}$ (Definition 4.6.2.11). In this case, the inclusion map $\iota : \operatorname{\mathcal{D}}_0 \hookrightarrow \operatorname{\mathcal{D}}$ is always a monomorphism in $\operatorname{\mathcal{QC}}$ (Corollary 9.2.4.33), and $F$ factors (uniquely) as the composition of $\iota$ with a functor $F_0: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}_0$. Beware that this factorization generally does not exhibit $\operatorname{\mathcal{D}}_0$ as an image of $F$ in the $\infty$-category $\operatorname{\mathcal{QC}}$, in the sense of Definition 10.2.3.1: that is, the functor $F_0$ need not be a quotient morphism in $\operatorname{\mathcal{QC}}$. This fails, for example, if $F$ is the inclusion functor $\operatorname{\partial \Delta }^{1} \hookrightarrow \Delta ^1$.

We now show that if $f: X \rightarrow Y$ is a morphism in $\operatorname{\mathcal{C}}$ which admits an image $Y_0$, then the subobject $Y_0 \subseteq Y$ is uniquely determined up to isomorphism.

Lemma 10.2.3.9. Let $q: X \twoheadrightarrow Y$ be a quotient morphism in an $\infty$-category $\operatorname{\mathcal{C}}$. Then every subterminal object $C \in \operatorname{\mathcal{C}}$ is $q$-local.

Proof. We wish to show that the composition map $\operatorname{Hom}_{\operatorname{\mathcal{C}}}(Y, C) \xrightarrow { \circ [q] } \operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,C)$ is a homotopy equivalence of Kan complexes. Since $C$ is subterminal, both mapping spaces are either empty or contractible. It will therefore suffice to show that if $\operatorname{Hom}_{\operatorname{\mathcal{C}}}(X,C)$ is nonempty, then $\operatorname{Hom}_{\operatorname{\mathcal{C}}}(Y,C)$ is also nonempty.

Let $\operatorname{\mathcal{C}}^{0}_{/Y}$ be the sieve generated by $q$. Since $q$ is a quotient morphism, $Y$ is a colimit of the diagram

$F: \operatorname{\mathcal{C}}^{0}_{/Y} \hookrightarrow \operatorname{\mathcal{C}}_{/Y} \rightarrow \operatorname{\mathcal{C}};$

that is, it can be lifted to an initial object $\widetilde{Y}$ of the coslice $\infty$-category $\operatorname{\mathcal{C}}_{F/}$. Since $C \in \operatorname{\mathcal{C}}$ is subterminal, the projection map $U: \operatorname{\mathcal{C}}_{/C} \rightarrow \operatorname{\mathcal{C}}$ restricts to a trivial Kan fibration from $\operatorname{\mathcal{C}}_{/C}$ to a sieve $\operatorname{\mathcal{C}}^{1} \subseteq \operatorname{\mathcal{C}}$. The assumption that $\operatorname{Hom}_{\operatorname{\mathcal{C}}}(X, C)$ is nonempty guarantees that $F$ takes values in $\operatorname{\mathcal{C}}^{1}$ and therefore factors through $\operatorname{\mathcal{C}}_{/C}$. A choice of factorization determines a lift of $C$ to an object $\widetilde{C} \in \operatorname{\mathcal{C}}_{F/}$. Since $\widetilde{Y}$ is an initial object of $\operatorname{\mathcal{C}}_{F/}$, we can choose a morphism $\widetilde{u}: \widetilde{Y} \rightarrow \widetilde{C}$ in the $\infty$-category $\operatorname{\mathcal{C}}_{F/}$. Applying the forgetful functor $\operatorname{\mathcal{C}}_{F/} \rightarrow \operatorname{\mathcal{C}}$, we obtain a morphism $u: Y \rightarrow C$ in $\operatorname{\mathcal{C}}$. $\square$

Lemma 10.2.3.10. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category and let $q: X \twoheadrightarrow Y$ be a quotient morphism in $\operatorname{\mathcal{C}}$. Then $q$ is left orthogonal to every monomorphism $i: C \hookrightarrow D$ of $\operatorname{\mathcal{C}}$.

Proof. Let $U: \operatorname{\mathcal{C}}_{/D} \rightarrow \operatorname{\mathcal{C}}$ denote the projection map, so that the monomorphism $i$ can be identified with a subterminal object $\widetilde{C} \in \operatorname{\mathcal{C}}_{/D}$ satisfying $U( \widetilde{C} ) = C$ (Remark 9.2.4.16). By virtue of Corollary 9.1.7.13, it will suffice to show that the object $\widetilde{C}$ is $\widetilde{q}$-local for every morphism $\widetilde{q}$ of $\operatorname{\mathcal{C}}_{/D}$ satisfying $U( \widetilde{q} ) = q$. This follows from Lemma 10.2.3.9, since $\widetilde{q}$ is a quotient morphism in the $\infty$-category $\operatorname{\mathcal{C}}_{/D}$ (Proposition 10.2.2.14). $\square$

Proposition 10.2.3.11. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category containing a $2$-simplex

10.23
$$\begin{gathered}\label{equation:universal-property-of-image} \xymatrix@R =50pt@C=50pt{ X \ar [d]^{q} \ar [dr]^{f} & \\ Y_0 \ar [r]^-{i_0} & Y } \end{gathered}$$

which exhibits $Y_0$ as an image of $f$. Then, for any monomorphism $i_1: Y_1 \hookrightarrow Y$ of $\operatorname{\mathcal{C}}$, the following conditions are equivalent:

$(1)$

The morphism $f$ factors (up to homotopy) through $i_1$. That is, there exists a $2$-simplex

10.24
$$\begin{gathered}\label{equation:universal-property-of-image2} \xymatrix@R =50pt@C=50pt{ X \ar [r]^-{g} \ar [dr]^{f} & Y_1 \ar [d]^{i_1} \\ & Y } \end{gathered}$$

in the $\infty$-category $\operatorname{\mathcal{C}}$.

$(2)$

The containment $[Y_0] \subseteq [Y_1]$ holds (where we regard the isomorphism classes $[Y_0]$ and $[Y_1]$ as elements of the partiallly ordered set $\operatorname{Sub}(Y)$; see Notation 9.2.4.26).

Proof. The implication $(2) \Rightarrow (1)$ follows immediately from the definitions. To prove the converse, we note that in the situation of $(1)$, we can amalgamate the diagrams (10.23) and (10.24) to obtain a lifting problem

$\xymatrix@R =50pt@C=50pt{ X \ar [d]^{q} \ar [r]^-{g} & Y_1 \ar [d]^{i_1} \\ Y_0 \ar [r]^-{i_0} \ar@ {-->}[ur] & Y }$

in the $\infty$-category $\operatorname{\mathcal{C}}$. Since $q$ is a quotient morphism and $i_1$ is a monomorphism, Lemma 10.2.3.10 guarantees that this lifting problem admits an (essentially unique) solution, which proves $(2)$. $\square$

Notation 10.2.3.12. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category and let $f: X \rightarrow Y$ be a morphism in $\operatorname{\mathcal{C}}$ which admits an image $Y_0 \subseteq Y$. It follows from Proposition 10.2.3.11 that the isomorphism class $[Y_0]$ is uniquely determined by $f$ (as an object of the partially ordered set $\operatorname{Sub}(Y)$; see Notation 9.2.4.26). To emphasize this, we will denote the isomorphism class $[Y_0]$ by $\operatorname{im}(f)$ and refer to it as the image of $f$. We will sometimes abuse notation by identifying $\operatorname{im}(f)$ with the object $Y_0$, viewed either as an object of the slice $\infty$-category $\operatorname{\mathcal{C}}_{/Y}$ or as an object of the $\infty$-category $\operatorname{\mathcal{C}}$.

Corollary 10.2.3.13. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category and let $f: X \rightarrow Y$ be a morphism of $\operatorname{\mathcal{C}}$. Then $f$ is an isomorphism if and only if it is both a monomorphism and a quotient morphism.

Proof. Without loss of generality, we may assume that $f$ is a monomorphism. Applying Example 10.2.3.5, we see that the isomorphism class $[X] \in \operatorname{Sub}(Y)$ is an image of $f$. It follows that $f$ is an isomorphism if and only if $\operatorname{im}(f) = [Y]$. By virtue of Proposition 10.2.3.6, this is equivalent to the requirement that $f$ is a quotient morphism. $\square$

Proposition 10.2.3.14 (Uniqueness of Images). Let $\operatorname{\mathcal{C}}$ be an $\infty$-category and let $\operatorname{Fun}'( \Delta ^2, \operatorname{\mathcal{C}})$ denote the full subcategory of $\operatorname{Fun}( \Delta ^2, \operatorname{\mathcal{C}})$ spanned by those $2$-simplices

$\xymatrix@R =50pt@C=50pt{ & Y_0 \ar [dr]^{i} & \\ X \ar [ur]^{q} \ar [rr]^{f} & & Y }$

which exhibit $Y_0$ as an image of $f$ (see Remark 10.2.3.3). Then the restriction functor

$D: \operatorname{Fun}'( \Delta ^2, \operatorname{\mathcal{C}}) \rightarrow \operatorname{Fun}( \Delta ^1, \operatorname{\mathcal{C}}) \quad \quad \sigma \mapsto d^{2}_{1}( \sigma )$

is a trivial Kan fibration from $\operatorname{Fun}'(\Delta ^2, \operatorname{\mathcal{C}})$ to the full subcategory $\operatorname{Fun}'( \Delta ^1, \operatorname{\mathcal{C}}) \subseteq \operatorname{Fun}( \Delta ^1, \operatorname{\mathcal{C}})$ spanned by those morphisms $f: X \rightarrow Y$ which admit an image in $\operatorname{\mathcal{C}}$.

Proof. Combining Lemma 10.2.3.10 with Theorem 9.1.8.2, we deduce that the functor $D$ is fully faithful, and therefore induces an equivalence from $\operatorname{Fun}'( \Delta ^2, \operatorname{\mathcal{C}})$ to the full subcategory $\operatorname{Fun}'(\Delta ^1, \operatorname{\mathcal{C}}) \subseteq \operatorname{Fun}( \Delta ^1, \operatorname{\mathcal{C}})$. To complete the proof, it will suffice to show that $D$ is an isofibration (Proposition 4.5.5.20). This follows from Corollary 4.4.5.3, since $\operatorname{Fun}'( \Delta ^2, \operatorname{\mathcal{C}})$ is a replete subcategory of $\operatorname{Fun}( \Delta ^2, \operatorname{\mathcal{C}})$ (see Corollary 10.2.2.12 and Remark 9.2.4.24). $\square$

Remark 10.2.3.15. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category and let $f: X \rightarrow Y$ be a morphism of $\operatorname{\mathcal{C}}$. Suppose that $\operatorname{\mathcal{C}}$ admits fiber products, so that $f$ admits a Čech nerve $\operatorname{\check{C}}_{\bullet }( X/Y )$ (Proposition 10.1.5.6). If $f$ has an image, then $\operatorname{im}(f)$ can be identified with the geometric realization of the underlying simplicial object of $\operatorname{\check{C}}_{\bullet }(X/Y)$. To see this, choose a $2$-simplex

$\xymatrix@R =50pt@C=50pt{ & Y_0 \ar [dr]^{i} & \\ X \ar [ur]^{q} \ar [rr]^{f} & & Y }$

which exhibits $Y_0$ as an image of $f$. Since $i$ is a monomorphism, Remark 10.2.1.17 supplies an isomorphism between the underlying simplicial objects of $\operatorname{\check{C}}_{\bullet }(X/Y)$ and $\operatorname{\check{C}}_{\bullet }(X/Y_0)$. It will therefore suffice to show that $\operatorname{\check{C}}_{\bullet }(X/Y_0)$ is a colimit diagram in $\operatorname{\mathcal{C}}$, which is a reformulation of our assumption that $q$ is a quotient morphism (Proposition 10.2.2.4).

Warning 10.2.3.16. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category which admits fiber products, let $f: X \rightarrow Y$ be a morphism of $\operatorname{\mathcal{C}}$, and let $C_{\bullet }$ denote the underlying simplicial object of the Čech nerve $\operatorname{\check{C}}_{\bullet }(X/Y)$. Remark 10.2.3.15 asserts that if $f$ has an image, then that image can be identified with a geometric realization of $C_{\bullet }$. Beware that the converse is false in general. Suppose that $C_{\bullet }$ admits a geometric realization $| C_{\bullet } |$, given by the image on an initial object $E$ of the coslice $\infty$-category $\operatorname{\mathcal{C}}_{ C_{\bullet } / }$. The augmented simplicial object $\operatorname{\check{C}}_{\bullet }(X/Y)$ determines another object $\widetilde{Y} \in \operatorname{\mathcal{C}}_{ C_{\bullet } / }$, so there is an (essentially unique) morphism from $E$ to $\widetilde{Y}$. The forgetful functor $\operatorname{\mathcal{C}}_{C_{\bullet } / } \rightarrow \operatorname{\mathcal{C}}_{X/}$ carries this morphism to a $2$-simplex

10.25
$$\begin{gathered}\label{equation:realization-not-image} \xymatrix@C =50pt@R=50pt{ & | C_{\bullet } | \ar [dr]^{i} & \\ X \ar [ur]^{q} \ar [rr]^{f} & & Y } \end{gathered}$$

in the $\infty$-category $\operatorname{\mathcal{C}}$. In this situation, the following conditions are equivalent:

• The morphism $i$ is a monomorphism.

• The diagram (10.25) exhibits $| C_{\bullet } |$ as an image of $f$.

• The morphism $f$ has an image in $\operatorname{\mathcal{C}}$.

Definition 10.2.3.17. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category. We say that $\operatorname{\mathcal{C}}$ has images if every morphism $f: X \rightarrow Y$ of $\operatorname{\mathcal{C}}$ has an image $\operatorname{im}(f) \in \operatorname{Sub}(Y)$.

Remark 10.2.3.18 (Functoriality of Images). Let $\operatorname{\mathcal{C}}$ be an $\infty$-category and let $\operatorname{Fun}'( \Delta ^2, \operatorname{\mathcal{C}}) \subseteq \operatorname{Fun}( \Delta ^2, \operatorname{\mathcal{C}})$ be the full subcategory described in Proposition 10.2.3.14. Then $\operatorname{\mathcal{C}}$ has images if and only if the restriction functor

$D: \operatorname{Fun}'( \Delta ^2, \operatorname{\mathcal{C}}) \rightarrow \operatorname{Fun}( \Delta ^1, \operatorname{\mathcal{C}}) \quad \quad \sigma \mapsto d^{2}_{1}( \sigma )$

is a trivial Kan fibration. If this condition is satisfied, then $D$ admits a section which carries each morphism $f: X \rightarrow Y$ of $\operatorname{\mathcal{C}}$ to a $2$-simplex

$\xymatrix@R =50pt@C=50pt{ & \operatorname{im}(f) \ar [dr] & \\ X \ar [ur] \ar [rr]^{f} & & Y }$

which exhibits $\operatorname{im}(f)$ as an image of $f$. In particular, we can promote the construction $f \mapsto \operatorname{im}(f)$ as a functor of $\infty$-categories $\operatorname{Fun}(\Delta ^1,\operatorname{\mathcal{C}}) \rightarrow \operatorname{\mathcal{C}}$.

Remark 10.2.3.19. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category, let $Q$ denote the collection of all quotient morphisms in $\operatorname{\mathcal{C}}$, and let $M$ denote the collection of all monomorphisms in $\operatorname{\mathcal{C}}$. Then $Q$ and $M$ are closed under isomorphism (see Corollary 10.2.2.12 and Remark 9.2.4.24), and $Q$ is left orthogonal to $M$ (Lemma 10.2.3.10). It follows that $\operatorname{\mathcal{C}}$ has images if and only if the pair $(Q,M)$ is a factorization system on $\operatorname{\mathcal{C}}$ (Definition 9.1.9.1).

Proposition 10.2.3.20. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category and let $f: X \rightarrow Y$ be a morphism of $\operatorname{\mathcal{C}}$. If $\operatorname{\mathcal{C}}$ has images, the following conditions are equivalent:

$(1)$

The morphism $f$ is a quotient morphism.

$(2)$

The morphism $f$ is left orthogonal to every monomorphism in $\operatorname{\mathcal{C}}$.

$(3)$

The morphism $f$ is weakly left orthogonal to every monomorphism in $\operatorname{\mathcal{C}}$.

Corollary 10.2.3.21. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category which has images, and let

$\xymatrix@R =50pt@C=50pt{ & Y \ar [dr]^{g} & \\ X \ar [ur]^{f} \ar [rr]^{h} & & Z }$

be a $2$-simplex of $\operatorname{\mathcal{C}}$, where $f$ is a quotient morphism. Then $g$ is a quotient morphism if and only if $h$ is a quotient morphism. In particular, the collection of quotient morphisms is closed under composition.

Corollary 10.2.3.22. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category containing a pushout diagram

$\xymatrix@R =50pt@C=50pt{ X \ar [d]^{f} \ar [r] & X' \ar [d]^{f'} \\ Y \ar [r] & Y'. }$

If $\operatorname{\mathcal{C}}$ is has images and $f$ is a quotient morphism, then $f'$ is also a quotient morphism.

Corollary 10.2.3.23. Let $\operatorname{\mathcal{C}}$ be an $\infty$-category with images. Then the collection of quotient morphisms in $\operatorname{\mathcal{C}}$ is closed under retracts (in the $\infty$-category $\operatorname{Fun}(\Delta ^1, \operatorname{\mathcal{C}})$).

Exercise 10.2.3.24. Show that the conclusion of Corollary 10.2.3.23 holds for every $\infty$-category $\operatorname{\mathcal{C}}$: that is, it is not necessary to assume that $\operatorname{\mathcal{C}}$ has images.