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This question already has an answer here:

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.

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marked as duplicate by Eric Wofsey category-theory Jun 24 at 22:21

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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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