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.


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 $ih=j$.

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    $\begingroup$ +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. $\endgroup$ – Noah Schweber Aug 24 '16 at 4:42
  • $\begingroup$ 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. $\endgroup$ – Lonatico Aug 24 '16 at 13:00
  • $\begingroup$ Right, the image isn't even a category at all. $\endgroup$ – Eric Wofsey Aug 24 '16 at 17:50
  • $\begingroup$ Nice. Thank you Eric for the complete explanation, and you too Noah for pointing the inejctive part. $\endgroup$ – Lonatico Aug 24 '16 at 21:49

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