$F(f)=0$, then $f=0$ where $0$ is the zero morphism for equivalent functors.

Question

For an additive category $\mathcal{C}$ and let $F:\mathcal{C} \to \mathcal{C}$ be a equivalent additive and covariant functor. Prove that for every morphsim $f:A \to B$ such $F(f)=0$, then $f=0$ where $0$ is the zero morphism in $Hom_{\mathcal{C}}(A,B)$.

My attemp to prove this goes as follow: As $F:\mathcal{C} \to \mathcal{C}$ be a equivalent functor we got that $F$ is full, faithful and dense by:

Fully faithful and essentially surjective is an equivalence

Now as $F(0)=0=F(f)$ we have to somehow use that $F$ is full by previous fact in order to show $f=0$.

When you say "an equivalent functor" does it mean that it is an equivalence of category ? — Captain Lama, Apr 20 '19 at 23:30

score 1 · Accepted Answer · answered Apr 20 '19 at 23:33

1

Since $F$ is faithful, it is by definition injective on morphism, i.e. the map $F_{A,B}:Hom_C(A,B)\to Hom_C(F(A),F(B))$ is injective. That means that if $F(f):F(A)\to F(B)$ is the zero morphism, since $F(0)=0$ by additivity, then $f:A\to B$ must be zero as well.

answered Apr 20 '19 at 23:33

Javi

6,541

¡Gracias! @Javi – Cos Apr 21 '19 at 04:00
De nada :) @Cos – Javi Apr 21 '19 at 11:21

$F(f)=0$, then $f=0$ where $0$ is the zero morphism for equivalent functors.

1 Answers1