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The analogy between functions from a set to set, and functors from a category to a category is obvious. But transferring terminology from functions to functors can be a tricky business.

I would like to ask if the following is acceptable: For $B$ a subcategory of $A$, and $D$ a subcategory of $C$, and $F:A \to C$ a functor, is the following statement clear in meaning, and well-defined?

It holds that the image of $B$ under $F$ lies in $D$.

If not, then how should it be phrased?

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  • $\begingroup$ Presumably you mean that the objects $F(b)$ for $b$ an object of $B$ are all objects of $D$. Do you mean the same for the morphisms? $\endgroup$
    – Jonathan Beardsley
    Commented Sep 19, 2012 at 13:54
  • $\begingroup$ Yes in both cases. $\endgroup$
    – Mihail Matrix
    Commented Sep 19, 2012 at 13:59
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    $\begingroup$ Analogy? A function is a functor (between discrete categories). $\endgroup$
    – Qiaochu Yuan
    Commented Sep 19, 2012 at 16:15
  • $\begingroup$ It seems still unclear. @Mihail: May I ask you to clarify if the intended meaning of your statement is the one explained by Andreas and Todd, or the one suggested by me? $\endgroup$
    – Fred Rohrer
    Commented Sep 19, 2012 at 20:40
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    $\begingroup$ @Theo: Of course. The point is the distinction between "is" and "is equivalent to". For example, the last "are" in your comment is of the latter type (supposing our definitions of "functor" coincide, and supposing we have a similar set theory). And in that case it is indeed appropriate to speak of analogy. $\endgroup$
    – Fred Rohrer
    Commented Sep 20, 2012 at 5:13

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If you use James Cranch's suggestion, namely to say that the image is a subcategory of $D$, you should make sure that it's actually true in your situation, because it might not be in general. $B$ might contain morphisms $f:x\to y$ and $g:z\to w$, and your functor might have $F(y)=F(z)$, so that $F(f)$ and $F(g)$ are composable morphisms in $D$. Yet, if $y\neq z$ and so $f$ and $g$ are not composable in $A$, then the composite $F(g)F(f)$ might not be in the image of $F$.

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Andreas's answer is very much to the point. I'd like to add to it and give some accepted terminology.

As he was saying, the naive image doesn't work. Consider the arrow category $2 = \{0 \to 1 \}$, and take the functor $2 \to Set$ which maps both objects to the natural numbers $\mathbb{N}$ and the arrow to the successor function $s: \mathbb{N} \to \mathbb{N}$. Now ask yourself if the "image" of this functor is actually a subcategory!

The notion you really want is called the essential image.

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As proven by Jon's comment, the statement is not clear in meaning. A precise statement is the following, but it is of course unclear whether it is equivalent to the statement you had in mind:

$F$ induces by restriction and coastriction a functor from $B$ to $D$.

(The very useful word "coastriction" seems to be not so well-known, but it is explained in this answer.)

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I think this is well-defined. You mean that the images of objects and morphisms in $B$ under $F$ all lie within $D$.

It might be slightly clearer to say, "... is a subcategory of $D$" rather than "... lies in $D$" (to avoid a reader thinking that the image has to be a single point for some stupid reason) but this is almost pedantic.

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