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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.

Anonymous Coward
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    Now I see what I was missing is that morphisms previously unable to composite can composite after the functor. – Anonymous Coward Jun 06 '13 at 16:43
  • I don't understand this. The "naive image" is just a collection of objects, and therefore can be regarded as a full subcategory. You can only restrict to morphisms which are induced by the functor. But again this is a subcategory. – Martin Brandenburg Jun 07 '13 at 10:17
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    @MartinBrandenburg: the concept you are describing seems to be called "full image" on nCatLab: http://ncatlab.org/nlab/show/full+image. – Blaisorblade Sep 17 '14 at 00:50

2 Answers2

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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.

Ittay Weiss
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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.

Dennis
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