# Can a functor $F: A\to B$ not map $A$ to a subcategory of $B$? [duplicate]

So, I'm studying Category Theory, and I'm stuck with some questions. One of them is this one:

Can a functor $$F:A \to B$$, meaning, a functor $$F$$ from category $$A$$ to category $$B$$, have an image that is NOT a subcategory of $$B$$?

I can't really find a situation where this happens, but by my readings this is wrong of my part.

Thanks in advance.

## 1 Answer

Yes, this can happen. The trick is that there might be morphisms which are not composable in $A$, but their image in $B$ is composable, and so the image of $F$ need not contain the composition of their images.

Here's a specific example. Let $A$ have four objects objects $a,b,c$ and $d$, with only two morphisms besides the identity, a morphism $f:a\to b$ and a morphism $g:c\to d$. Let $B$ have three objects $x,y,z$ with non-identity maps $h:x\to y$, $i:y\to z$, and $j:x\to z$, with $j=ih$. Define a functor $F:A\to B$ by $F(a)=x$, $F(b)=F(c)=y$, $F(d)=z$, $F(f)=h$, and $F(g)=i$. Then the image of $F$ is not a subcategory, because it contains $h$ and $i$ but not $ih=j$.

• +1. Note for the OP that if $F$ is injective on objects, then this can't happen and the image of $F$ is indeed a subcategory. – Noah Schweber Aug 24 '16 at 4:42
• Ok, but if the image does not contain ih, would it be a category at all? I mean, as far as I understand, a category must have the compositions of its morfisms. – Lonatico Aug 24 '16 at 13:00
• Right, the image isn't even a category at all. – Eric Wofsey Aug 24 '16 at 17:50
• Nice. Thank you Eric for the complete explanation, and you too Noah for pointing the inejctive part. – Lonatico Aug 24 '16 at 21:49