Above the identity there must lie an isomorphism.

Let $F:\mathcal C\to \mathcal D$ be a (covariant) functor between two categories. Of course, a functor sends isomorphisms to isomorphisms. And "over" an isomorphism in $\mathcal D$ there might be an arrow which is not an isomorphism. That is, not every functor reflects isomorphisms.

But in a paper, it seems to be obvious that over the identity $D\overset{1_D}{\longrightarrow} D$ of an object in $\mathcal D$, that is, in $$\{g\in \textrm{mor}_\mathcal C\,|\,F(g)=1_D\},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(\star)$$ everything is an isomorphism. Why is this true? Does it hold a priori for other morphisms rather than the identities $1_D$?

Motivation: I want to verify that the subcategory $\mathcal C_D\subset \mathcal C$ whose objects are objects of $\mathcal C$ that are sent to $D$ by $F$, and whose morphisms are those in $(\star)$, is a groupoid.

Thanks for any hint!

• The answer to this question depends upon the nature of your functor. If your functor is an equivalence for instance , then star is true. – Baby Dragon Feb 1 '13 at 17:10
• @Baby Dragon: Yes, you are right. It seemed to me that the categories involved were general enough, but I was wrong. – Brenin Feb 1 '13 at 18:55
• It is true for fibered categories. – Martin Brandenburg Feb 2 '13 at 14:33

It must be a consequence of the choice of functor $F$ and categories $\mathcal C$ and $\mathcal D$ because ($\ast$) is not true in general.
For a counterexample let $\mathcal{C = D}$ be the category with one object $\{\ast\}$ and endomorphisms $\mathrm{End}_{\mathcal C}(\ast)$ equal to the monoid $(\mathbb N_0, +)$, where $0 \in \mathbb N_0$ represents the identity map. Then a functor $F\colon \mathcal{C \to D}$ is equivalent to a monoid endomorphism of $\mathbb N_0$. If we chose $x \mapsto 0 \ \forall x$ then we get that every map in $\mathcal C$ maps to the identity map but the identity map is the only isomorphism in $\mathcal C$ so there are plenty of maps over the identity that are not isomorphisms.