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 \begin{equation} \{g\in \textrm{mor}_\mathcal C\,|\,F(g)=1_D\},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(\star) \end{equation} 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!

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

  • $\begingroup$ Curiosity: is every monoid a possible endomorphism set of a category with just one object? $\endgroup$ – Brenin Feb 1 '13 at 19:12
  • $\begingroup$ Yes, every monoid defines a category with one object whose endomorphisms are that monoid. Conversely every small category with one object defines a monoid by taking the endomorphisms of that object. So in fact you get an isomorphism between the category of all small categories with one object and the category of all monoids. $\endgroup$ – Jim Feb 1 '13 at 20:44
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    $\begingroup$ @Jim, strictly speaking, the category of all monoids and the category of all small categories with one object are equivalent categories, but are not isomorphic categories (because of the choice of the single object that must be made). $\endgroup$ – PatrickR Feb 19 '13 at 4:54
  • $\begingroup$ @PatrickR: Yep, thanks for pointing that out. I make that mistake rather often :) $\endgroup$ – Jim Feb 19 '13 at 4:58

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