# 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.

## marked as duplicate by Eric Wofsey category-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jun 24 at 22:21

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