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A functor $F$ is defined to be a mapping from category $\mathcal{C}$ to $\mathcal{D}$ such that:

(1) $F(f\circ_{\mathcal{C}} g)=F(f)\circ_{\mathcal{D}} F(g)$ (say, for a covariant functor).

(2) $F(id_{A})=id_{F(A)}$

Question: why is (2) necesssary?


Here is an attempt to show that (1) implies (2), I would appreciate to be told where it fails:

Take any $f\in hom_{\mathcal{C}}(A,B)$, $g\in hom_{\mathcal{C}}(B,A)$ Then $$ F(f)\circ F(id_{A}) \stackrel{(1)}{=} F(f\circ id_A) = F(f) $$ and $$ F(id_A) \circ F(g) \stackrel{(1)}{=} F(id_A \circ g) = F(g) $$

So that $F(id_A)$ satisfies the axioms of an identity (assuming that $F$ is surjective, though I'm not sure what that means for functors), and since identities are unique (an easy lemma) then $id_{F(A)}=F(id_A)$.

The two problems, as noted are:

(a) $F$ might not be surjective.

(b) $hom_{\mathcal{C}}(A,B)$ or $hom_{\mathcal{C}}(B,A)$ might be empty.

Can anyone explain more about this? Show an non-stupid example where something would fail to be a functor because of this?

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  • $\begingroup$ I believe what you are describing is called a prefunctor (fairly obscure.) $\endgroup$ Aug 31, 2015 at 0:49
  • $\begingroup$ A functor sends isomorphisms to isomorphisms; if you removed (2) this would no longer be the case. $\endgroup$ Aug 31, 2015 at 8:35
  • $\begingroup$ What about a "functor" from an abelian category to itself that is identity on objects, but replaces all the morphisms with the zero morphism. $\endgroup$ Sep 1, 2015 at 11:52

3 Answers 3

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Consider the following counterexample.

Let $\mathcal{C}$ be the category with one object $C$ and one morphism $\mathrm{id}_C$. Let $\mathcal{D}$ be the category with one object $D$ and two morphisms $\mathrm{id}_D : D \to D$ and $f : D \to D$, where composition is defined by $f \circ f = f$.

Define $F : \mathcal{C} \to \mathcal{D}$ by $$F(C)=D \qquad \text{and} \qquad F(\mathrm{id}_C)=f$$ Then $F$ satisfies all the requirements of being a functor, except for the requirement that $F(\mathrm{id}_C)=\mathrm{id}_{D}$, since $F(\mathrm{id}_C)=f\ne\mathrm{id}_D$.


Added: As per your question in the comments, here's an example where $F$ is surjective on morphisms: take $\mathcal{C}$ to be a category with two objects $C_1$ and $C_2$ and only identity morphisms, take $\mathcal{D}$ as before, define $F(C_1)=F(C_2)=D$, $F(\mathrm{id}_{C_1})=f$ and $F(\mathrm{id}_{C_2})=\mathrm{id}_D$.

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  • $\begingroup$ So this fails because $F$ is not "surjective"? The map $id_D$ is not covered. Is there also a surjective counter-example? $\endgroup$
    – PPR
    Aug 30, 2015 at 17:56
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    $\begingroup$ @PPR: See updated answer. $\endgroup$ Aug 30, 2015 at 18:05
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If $f:\cal C\to D$ is full, then (1) implies (2):
Let $x$ be an object of $\cal C$ and let $y=f(x)$. Assume $e$ is one of the arrows at $x$ such that $f(e)=1_y$. Then we have $$ f(1_x) = f(1_x)1_y = f(1_x)f(e) = f(1_xe) = f(e) = 1_y $$ Also, if $\cal D$ is a groupoid, (1) implies (2) since $$ f(1_x) = f(1_x)1_y = f(1_x)f(1_x)f(1_x)^{-1} f(1_x 1_x)f(1_x)^{-1} = f(1_x)f(1_x)^{-1} = 1_y $$

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If $M$ is a category with one object $O$, then $(\mathrm{Hom}(O, O), ∘)$ is a monoid, and conversly, every monoid $M$ can be made into a one-object category the morphisms of which are the elements of that monoid, and composition is monoid product. Furthermore, monoid morphisms correspond to functors between the one object categories constructed in this way. This is often a simple and useful example to keep in mind

Now take any semigroup morphism of two monoids that isn't a monoid morphism (ie. it satisfies $f(ab) = f(a)f(b)$, but not $f(e) = e$), and you'll get your counterexample. To construct these, it's easiest to note that the unit must map to an idempotent, and go from there: for example, $(ℕ, \cdot) → (ℕ, \cdot)$, $n ↦ 0$, or $(ℕ, \cdot) → (ℕ × ℕ, \cdot)$, $n ↦ (n, 0)$ will work.

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    $\begingroup$ This is a tripe, but... are you sure the "inclusion of even numbers" is an example? Depending on how you define $\mathbb{N}$, it is either a submonoid, or a subsemigroup of a non-monoid. $\endgroup$ Apr 21, 2016 at 18:43
  • $\begingroup$ @darijgrinberg: No, you are right of course, it should be correct now. $\endgroup$
    – user54748
    Apr 22, 2016 at 3:56

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