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Let's call a functor $T\colon\mathcal{C}\to\mathcal{D}$ imageable, if its image on objects and morphisms forms a subcategory in $\mathcal{D}$. More formally, a functor $T\colon\mathcal{C}\to\mathcal{D}$ is imageable iff there exists such subcategory $\mathcal{B}$ in $\mathcal{D}$, that $T_{\text{Obj}}(\text{Obj}(\mathcal{C}))=\text{Obj}(\mathcal{B})$ and $T_{\text{Mor}}(\text{Mor}(\mathcal{C}))=\text{Mor}(\mathcal{B})$.

My questions are:

  1. Is this property of functors well-known? (in particular, does it have a name?)
  2. Is there any "if and only if" condition for a functor $T$ to be imageable (of course, except of trivial reformulations)?
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    $\begingroup$ Every functor is imageable. $\endgroup$ – Alex Kruckman Mar 28 '16 at 18:15
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    $\begingroup$ Not every functor is imageable, because there might be maps in $\mathcal{C}$ that were not composable but become composable after applying $T$. $\endgroup$ – Eric Wofsey Mar 28 '16 at 18:18
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    $\begingroup$ @AlexKruckman No. Take a functor $T\colon\mathbf{2}\sqcup\mathbf{2}\to\mathbf{3}$, such that $T(0,0)=0$, $T(1,0)=1$, $T(0,1)=1$, $T(1,1)=2$. $\endgroup$ – Oskar Mar 28 '16 at 18:18
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    $\begingroup$ In any case, I have never heard of this condition, and it doesn't sound very natural. In particular, it is very "evil" (i.e., not invariant under equivalences of categories). $\endgroup$ – Eric Wofsey Mar 28 '16 at 18:21
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    $\begingroup$ @EricWofsey Good point, thanks. But I'm still interested in a possible solution. $\endgroup$ – Oskar Mar 28 '16 at 18:24
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I realize it is very evil to phrase things the way they are in this answer. At the same time, it seems that this question is inherently evil, so it seems like any answer to it must be evil as well.

Claim ($\implies$) If the image of a functor $T$ is not a category, then there exist objects $D_1, D_2, D_3 \in T_{\text{Obj}}(\text{Obj}(\mathcal{C}))$ and morphisms $g_1, g_2 \in T_{\text{Mor}}(\text{Mor}(\mathcal{C}))$, with $D_1 \overset{g_1}{\to} D_2$ and $D_2 \overset{g_2}{\to} D_3$, such that $g_2 \circ g_1 \not\in T_{\text{Mor}}(\text{Mor}(\mathcal{C}))$. (Clearly $g_2 \circ g_1$ exists and is an element of $\text{Mor}(\mathcal{D})$ since $\mathcal{D}$ is a category.)

Proof(?) of claim: The image of $T$ for any functor $T$ always has an obvious "set of objects", namely $T_{\text{Obj}}(\text{Obj}(\mathcal{C}))$, and "sets of morphisms", $T_{\text{Mor}}(\text{Mor}(\mathcal{C}))$. That means if the image of $T$ is not a (sub)category, then it must fail one of the category axioms:

  1. existence of identity morphisms
  2. given any morphisms $D_1 \overset{g_1}{\to} D_2$ and $D_2 \overset{g_2}{\to} D_3$, then there must exist a morphism $D_1 \overset{g_2 \circ g_1}{\to} D_3$,
  3. associativity, $g_3 \circ (g_2 \circ g_1) = (g_3 \circ g_2) \circ g_1$.

For any object $D \in T_{\text{Obj}}(\text{Obj}(\mathcal{C}))$, by definition there exists a $C \in \mathcal{C}$ such that $T(C) = D$, and by the functor axioms, one has that $T(\operatorname{Id}_C)= \operatorname{Id}_{T(C)} = \operatorname{Id}_D$, thus not only does $\operatorname{Id}_D$ exist in $\text{Mor}(\mathcal{D})$, it is the image of a morphism in $\text{Mor}(\mathcal{C})$ and thus in $T(\text{Mor}(\mathcal{C}))$, i.e. the image of any functor always satisfies 1.

Therefore, if the image of $T$ is not a (sub)category, then it must fail either 2. (closure under composition) or 3. (associativity). However, associativity is automatically inherited from $\mathcal{D}$. So if the image of a functor $T$ is not a subcategory, then it must fail axiom 2. (closure under composition).

The statement to be proved is just the negation of 2. So basically the above proof is just a string of proofs by contradiction, which is evil for reasons other than violating the principle of equivalence.

Claim ($\impliedby$) If there exist objects $D_1, D_2, D_3 \in T_{\text{Obj}}(\text{Obj}(\mathcal{C}))$ and morphisms $g_1, g_2 \in T_{\text{Mor}}(\text{Mor}(\mathcal{C}))$, with $D_1 \overset{g_1}{\to} D_2$ and $D_2 \overset{g_2}{\to} D_3$, such that $g_2 \circ g_1 \not\in T_{\text{Mor}}(\text{Mor}(\mathcal{C}))$, then the image of $T$ is not a category.

Proof: This is clear, since the assumption means that the image violates the closure under composition axiom. Basically every counterexample to the image of $T$ being a subcategory which I have found has used violating that axiom to show it's not a subcategory, see e.g. (1)(2)(3).


Therefore, regarding your question of finding interesting equivalent conditions for a functor to not be imageable, I believe one can reduce it to the question of finding interesting conditions equivalent to:

($*$) There exist objects $D_1, D_2, D_3 \in T_{\text{Obj}}(\text{Obj}(\mathcal{C}))$ and morphisms $g_1, g_2 \in T_{\text{Mor}}(\text{Mor}(\mathcal{C}))$, with $D_1 \overset{g_1}{\to} D_2$ and $D_2 \overset{g_2}{\to} D_3$, such that $g_2 \circ g_1 \not\in T_{\text{Mor}}(\text{Mor}(\mathcal{C}))$.

Clearly as far as conditions go, this is "sub-optimal", since it is in terms of objects and morphisms of $\mathcal{D}$, whereas the whole point of having a functor $T: \mathcal{C} \to \mathcal{D}$ is to be able to understand things in terms of objects and morphisms of $\mathcal{C}$. I make the following:

Claim: Condition ($*$) is equivalent to:
($**$) There exist in $\mathcal{C}$ morphisms $C_1 \overset{f_1}{\to} C_2$ and $C_3 \overset{f_2}{\to} C_4$ such that $C_2 \not=C_3$ (so $f_2 \circ f_1$ isn't defined) but $T(C_2) = T(C_3)$, and there does not exist in $\mathcal{C}$ any morphism $C_5 \overset{f_3}{\to} C_6$ such that $T(f_3) = T(f_2) \circ T(f_1)$.

Proof(?) $(**) \implies (*)$
Clearly $T(f_1), T(f_2) \in T_{\text{Mor}}(\text{Mor}(\mathcal{C}))$ and clearly $ T(f_2) \circ T(f_1) \not\in T_{\text{Mor}}(\text{Mor}(\mathcal{C}))$. So closure under composition is violated so the image of $T$ is not a category, which is equivalent to $(*)$.

Proof(?) $(*) \implies (**)$
$g_1, g_2 \in T_{\text{Mor}}(\text{Mor}(\mathcal{C}))$, with $D_1 \overset{g_1}{\to} D_2$ and $D_2 \overset{g_2}{\to} D_3$ if and only if there exist in $\mathcal{C}$ morphisms $C_1 \overset{f_1}{\to} C_2$ and $C_3 \overset{f_2}{\to} C_4$ such that $T(f_1) = g_1$ and $T(f_2) = g_2$. Likewise, $g_2 \circ g_1 \not\in T_{\text{Mor}}(\text{Mor}(\mathcal{C}))$ if and only if there does not exist in $\mathcal{C}$ any morphism $C_5 \overset{f_3}{\to} C_6$ such that $T(f_3) = g_2 \circ g_1 = T(f_2) \circ T(f_1)$.

Assume by means of contradiction that $C_2 = C_3$. Then $f_2 \circ f_1$ is defined, and so by properties of functors, one has that $T(f_2 \circ f_1) = T(f_2) \circ T(f_1) = g_2 \circ g_1$, thus $g_2 \circ g_1 \in T_{\text{Mor}}(\text{Mor}(\mathcal{C}))$, contradiction.


Looking at things this way, it is not surprising that the functor $T$ being injective on objects is sufficient for the image of $T$ to be a subcategory, nor is it surprising that counterexamples are usually given with very small categories $\mathcal{C}$ (so that the non-existence condition, that no $C_5 \overset{f_3}{\to} C_6$ with $T(f_3) = T(f_2) \circ T(f_1)$ may exist, can be easily verified) and with two arrows in $\mathcal{C}$ whose codomain and domain don't line up, get sent by $T$ to two arrows in $\mathcal{D}$ whose codomain and domain do line up. (Since it seems that, by means of contradiction, this is literally the only way the image can fail to be a subcategory.)

It would possibly be nice if someone could provide a direct proof of these facts though, instead of rely on questionable contradiction proofs the way this answer does.

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    $\begingroup$ You are right, the question is "evil", as Eric Wofsey pointed out some years ago :) Sinсe then, I think that there is no actual answer to this question. But I don't think it should remain unanswered, and since you worked out some general points of "imageability", I accept your answer. $\endgroup$ – Oskar Mar 6 at 22:55
  • $\begingroup$ Thank you! Yeah, as far as I could work out there was no simpler way to phrase $(**)$, even though the last part of the condition could almost always be impossible to verify for an infinite category, and would be to tedious (at best) for most finite categories. At the very least I found thinking of it that way helpful for understanding why the counterexamples one finds on the internet almost always all look the same or very similar, as I guess it seems I already mentioned in the answer. $\endgroup$ – hasManyStupidQuestions Mar 6 at 23:40

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