# Conservativity implies faithfulness

It looks to me that one can state the following:

Let $\mathcal{K}$ and $\mathcal{M}$ be abelian categories. Then a functor $T:\mathcal{K} \to \mathcal{M}$ preserving the terminal object and cokernels is faithful if and only if is conservative.

My proof of this fact, which might be wrong is the following: If $T$ is conservative, then $f = g$ if and only if $T(\text{CoKer}(f-g)) = 0$ if and only if $T(f)=T(g)$. The other implication is even easier because abelian categories are balanced.

If this is correct the following are also true:

• Let $\mathcal{K}$ and $\mathcal{M}$ be abelian categories. Then a functor $T:\mathcal{K} \to \mathcal{M}$ preserving the terminal object and kernels is faithful if and only if is conservative.
• Let $\mathcal{K}$ and $\mathcal{M}$ be abelian categories. Then a functor $T:\mathcal{K} \to \mathcal{M}$ preserving finite limits is conservative if and only if is faithful.
• A right adjoint between abelian categories is faithful if and only if is conservative.

Remark

In every proof is enough for $\mathcal{K}$ and $\mathcal{M}$ to be enriched over groups, to have a terminal object and (co)kernels.

Thus here comes my question.

Q: Do you know other theorems like

• hypoteses... then conservative functors are precisely faithful ones.
• hypoteses... then a conservative functor is always faithful.
• By enriched over groups, I guess you mean "enriched over abelian groups", i.e. preadditive? The category of groups does not have a symmetric monoidal closed structure. – Arnaud D. Oct 9 '17 at 8:33
• Yes. You are right. – Ivan Di Liberti Oct 9 '17 at 8:33
• I've found this question, which is highly related to yours : math.stackexchange.com/questions/1513847/… – Arnaud D. Oct 9 '17 at 8:57

The cokernel of a zero map is an isomorphism, not zero, but the claim is still true: $Tf=Tg$ if and only if $T$ sends a cokernel map of $f-g$ to an isomorphism, if and only if the cokernel of $f-g$ is an isomorphism, if and only if $f=g$. The same argument works in the non-additive case when we observed that two maps are equal also just when their (co)equalizer is an isomorphism.