1
$\begingroup$

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.
$\endgroup$
3
  • $\begingroup$ 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. $\endgroup$ – Arnaud D. Oct 9 '17 at 8:33
  • $\begingroup$ Yes. You are right. $\endgroup$ – Ivan Di Liberti Oct 9 '17 at 8:33
  • $\begingroup$ I've found this question, which is highly related to yours : math.stackexchange.com/questions/1513847/… $\endgroup$ – Arnaud D. Oct 9 '17 at 8:57
1
$\begingroup$

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.

Entering the realm in which limits are weak, this is also true for exact functors of triangulated categories, because again zero maps are exactly those with isomorphic cones. A similar argument can be made in a homotopy category which is not triangulated for functors which preserve homotopy equalizers or coequalizers, since these are still defined up to non-canonical isomorphism. So, one sees there can't be a functor from a non-concrete homotopy category to sets, sending homotopy equalizers to equalizers, or dually.

$\endgroup$
3
  • $\begingroup$ Sure, thanks for the correction, I have been lazy in the proof. That was the idea. Thanks for the contributions. I did not accept because maybe someone else has other ideas, those are very close to the one I exposed. $\endgroup$ – Ivan Di Liberti Oct 8 '17 at 17:04
  • 1
    $\begingroup$ I don't understand what you're saying about pullbacks. The pullback of a map with itself is its kernel pair, which is in general not equal to the square of the domain... $\endgroup$ – Arnaud D. Oct 9 '17 at 7:39
  • 1
    $\begingroup$ @ArnaudD. Thanks, you're right. And anyway, a functor preserving pullbacks and the terminal object, as my construction would have required, already preserves equalizers in the first place. $\endgroup$ – Kevin Arlin Oct 9 '17 at 13:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.