The Image of a Groupoid Morphism is Not a Subgroupoid

Question

I am reading Brown's "Topology and Groupoids" book. In it he defines a subgroupoid as a subcategory of a groupoid that contains all the inverses for all the morphisms. A morphism of groupoids is a functor of categories. He claims that the image of a morphism of groupoids is not a subgroupoid. But, I believe it is a subgroupoid: If $F: G\rightarrow H$ is a morphism of groupoids, then $F(fg)=F(f)F(g)$ for all morphisms $f$ and $g$ in $G$. So, if $f$ and $g$ are inverses, so are $F(f)$ and $F(g)$. Am I missing something?

Answer 1

The issue is that that the image might not be a subcategory at all. It might be the case that $Ff$ and $Fg$ are composable in $H$ even though $f$ and $g$ are not composable in $G$. This means that $FfFg$ might not lie in the image of $F$.

Browns example of this is the functor $I \to \Bbb Z$, where $I$ is the groupoid $* \leftrightarrow *$, taking the morphism $* \rightarrow *$ to $1$ and the morphism $* \leftarrow *$ to $-1$. The image of this map of groupoids is the elements $\{0, \pm 1\}\in \Bbb Z$, which is not a subgroupoid.