Can it happen that the image of a functor is not a category?

Question

On Hilton and Stammbach's homological algebra book, end of chap. 2, they wrote in general $F(\mathfrak{C})$ is not a category at all in general. But I don't quite get it. I checked the axioms of a category for the image, and I think they are all satisfied. Am I missing something? Thanks.

Answer 1

Consider the category $C$ with four objects, $a,b,c,d$ and, other than identity arrows, a single arrow $a\to b$ and a single arrow $c\to d$. Now consider the category $D$ with three objects $x,y,z$, and, aside from identity arrows, the arrows $x\to y$, $y\to z$, and $x\to z$. Now, consider the functor $F:C\to D$ with $F(a)=x$, $F(b)=F(c)=y$, and $F(d)=z$ (extended uniquely to arrows). Its image is not a category.

This business is related to the fact that epis in $Cat$ are not so simple at all. In work of Isbell epis in $Cat$ are characterized. It's worth noting that regular epis, split epis, etc. in $Cat$ are quite different, attesting again to the subtlety of epis.

Answer 2

Drawing a picture of Ittay Weiss's answer:

\begin{align*} a \underset{f}{\longrightarrow}\, &b \\ &c \underset{g}{\longrightarrow} d \\ \\ &\Big{\downarrow}F \\ \\ x \underset{F(f)}{\longrightarrow}\,&y \underset{F(g)}{\longrightarrow} z \end{align*}

Here, the objects of the image are $x$, $y$, and $z$, and the arrows are $F(f)$ and $F(g)$. Categories are closed under composition of arrows, but $F(g)F(f)$ is not in the image of $F$ (there are no arrows $a \to d$), so the image of $F$ cannot be a category.

Answer 3

There are two questions:

1. Is the image of a functor always a category?

The answer is Yes.

2. If $F : \mathcal{C} \to \mathcal{D}$ is a functor, is the set of objects $F(\mathrm{Ob}(\mathcal{C}))$ together with the hom-sets $\hom(u,v) = \{F(f) : x,y \in \mathrm{Ob}(\mathcal{C}), \,u = F(x), \, v = F(y), f \in \hom(x,y)\}$ a subcategory of $\mathcal{D}$?

The answer is No.

In particular since we are in the context of category theory, we should use the notion of the image of a morphism and not apply any ad-hoc construction which typically ignores some structure (see my remark about forgetful functors below). We may apply this notion to the category of all categories $\mathbf{Cat}$ as well*, and see that the image of a functor $F : \mathcal{C} \to \mathcal{D}$ is the smallest subcategory of $\mathcal{D}$ through which $F$ factors. It always exists, and by definition it is a subcategory of $\mathcal{D}$.

Specifically, the objects of $\mathrm{im}(F)$ are the objects of the form $F(x)$ for $x \in \mathrm{Ob}(\mathcal{C})$, and a morphism $u \to v$ is a composition of morphisms $ u = u_1 \to u_2 \to \cdots \to u_n = v$, where each $u_i \to u_{i+1}$ has the form $F(f) : F(x) \to F(y)$ for some morphism $f : x \to y$ in $\mathcal{C}$.

This answers the first question. Ittay's example answers the second question, and it shows that setting $n=1$ above is not enough. The functor $F : \mathcal{C} \to \mathcal{D}$ in Ittay's example actually satisfies $\mathrm{im}(F) = \mathcal{D}$, and it is an epimorphism, so that it is an extremal epimorphism.

Somewhat similar, notice that the category $\mathbf{CompMet}$ of complete metric spaces has images, but they should not be confused with the images of the underlying maps of sets, because the forgetful functor $U : \mathbf{CompMet} \to \mathbf{Set}$ does not preserve images. Likewise, Ittay's example shows that the forgetful functor $\mathbf{Cat} \to \mathbf{Quiv}$ to the category of quivers (that forgets identities and compositions) does not preserve images, because what is written in 2. above is exactly how to construct images in $\mathbf{Quiv}$.

*Here, $\mathcal{U}$ is a fixed Grothendieck universe, and $\mathbf{Cat}$ is a shorthand for $\mathbf{Cat}_U$, the category of all $\mathcal{U}$-categories. We cannot possibly form the category of all categories in a literal sense.

Answer 4

Take $\mathcal C=\mathbf{Grp}$ the category of groups, $\mathcal D=\mathbf{Set}$ the category of sets, and $T$ the forgetful functor

Cite Mathworld entry faithful functor:
the forgetful functor from the category of groups to the category of sets is faithful, but it identifies non-isomorphic groups having the same underlying set

• $G_0$ be the group on $\{a,b\},aa=bb=a,ab=ba=b$
• $G_1$ be the group on $\{a,b\},aa=bb=b,ab=ba=a$

Consider the group isomorphisms

$$ f:G_0\to(\mathbb Z/2,+)\quad a\mapsto0\quad b\mapsto1 $$ $$ g:(\mathbb Z/2,+)\to G_1\quad 0\mapsto b\quad 1\mapsto a $$ Then $$ T(f) : \{a,b\}\to\{0,1\}\quad T(f)(a)=0\quad T(f)(b)=1 $$ $$ T(g) : \{0,1\}\to\{a,b\}\quad T(g)(0)=b\quad T(g)(1)=a $$ as functions of sets.

In $\mathbf{Set}$ we can compose them:$$T(g)\circ T(f):\{a,b\}\to\{a,b\}\quad a\mapsto b\quad b\mapsto a$$ For $\operatorname{Im}T$ to be a category, this composite must again be a morphism in $\operatorname{Im}T$; but it is not equal to $T(h)$ for any group homomorphism $h$ because $h$ must send the identity of the group to itself.

So $\operatorname{Im}T$ is not a category.

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If the object function of $T$ is injective, then $\operatorname{Im}T$ is a category

Proof

Assume $T$ is injective on objects. Take morphisms in $\operatorname{Im}T$

$$ T(f):T(C)\to T(C'), \qquad T(g):T(C')\to T(C'') $$

(where the middle really is the same object since $T$ is injective on objects). Then the composite in $\mathcal D$ is

$$ T(g)\circ T(f) = T(g\circ f), $$

which is again in $\operatorname{Im}T$. Identities are there because $T(\mathrm{id}_C)=\mathrm{id}_{T(C)}$. Thus $\operatorname{Im}T$ is closed under identities and composition and is therefore a category.