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I'm a newbie at category theory and just started reading Emily Riehl's Category Theory in Context. I got to the definition of functors, which contains the following two axioms: if $F:C\to D$ is a functor between categories, then

  • For every composable pair of morphisms $f:x\to y$, $g:y\to z$ in $C$, $F(g\circ f)=Fg\circ Ff$
  • For every object $c\in C$, $F\left(1_c\right)=1_{Fc}$

I'm concerned about the second axiom. For a group homomorphism $f:G\to G'$, it's enough to say that $f$ respects the group operation, and it follows from there that $f\left(1_G\right)=1_{G'}$. On the other hand, it's not enough to say that a ring homomorphism $g:R\to R'$ respects both addition and multiplication, for there exist maps that are both additive and multiplicative that don't preserve the multiplicative identity.

I tried coming up with a counterexample with a category consisting of one object and whose morphisms are the elements of a ring; however, this creates two identity morphisms (the additive and multiplicative identity). Even worse, these morphisms aren't necessarily associative.

Are there any examples of almost-functors that obey the first axiom but not the second?

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    $\begingroup$ Take the ring example and forget about the addition. $\endgroup$
    – Zhen Lin
    May 18 at 7:26
  • $\begingroup$ Oh fish thank you for the fast reply @ZhenLin ! That seems so obvious in hindsight now... $\endgroup$
    – doobdood
    May 18 at 7:30

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Here is the simplest example I can think of.

Let $\mathcal{1}$ be the category with a single object (call it $\bullet$) and a single morphism, which is the identity morphism on $\bullet$.

Let $\mathcal{C}$ be the category with one object (call it $*$) and a morphism $f\colon *\to *$ such that $f\circ f = f$. The morphisms $f$ and the identity $\mathrm{id}_*$ are the only morphisms in this category.

Then we can define an "almost-functor" $A\colon \mathcal{1}\to\mathcal{C}$ that maps $\bullet$ to $*$ and maps $\mathrm{id}_\bullet$ to $f$.

This almost-functor obeys the composition of morphisms because $A(\mathrm{id}_\bullet\circ\mathrm{id}_\bullet) = f = A(\mathrm{id}_\bullet)\circ A(\mathrm{id}_\bullet)$. However, it does not obey the identity axiom because it maps an identity morphism to a non-identity morphism.

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    $\begingroup$ Thank you! I came up with the category of a single object, say $\star$, and morphisms given by integers $n:\star\to\star$. Then, the composition of two morphisms is given by integer multiplication, i.e. $n\circ m=nm:\star\to\star$. Then, take the endofunctor that maps $\star$ to itself and $n:\star\to\star$ to $2n:\star\to\star$. Now that I think about it, any monoid "induces" a category in this way, and any map monoid-homomorphism induces an "almost-functor". $\endgroup$
    – doobdood
    May 18 at 7:46

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