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I am trying to solve Popescu, Abelian Categories with Applications to Rings and Modules, exercise 4, page 68. Assume that we have abelian categories $A,B$ and an additive faithful functor $T:A\to B$. We need to show that if $$ T(a) \stackrel{T(f)}{\longrightarrow} T(b)\stackrel{T(g)}{\longrightarrow} T(c) $$ is exact, then $a \stackrel{f}{\longrightarrow} b \stackrel{g}{\longrightarrow} c$ is exact as well. Thanks to the answer to this question, I know how to check that faithful functors reflect kernels and cokernels, but this doesn't seem to be enough to claim that they reflect exact sequences in general. In particular, I have some problems in checking that $\ker(T(g))\cong T\left(\ker(g)\right)$.

Then I found Eilenberg, Moore, Foundations of Relative Homological Algebra, Proposition 1.2, in which they explicitly ask for $T$ being kernel preserving in order to prove the foregoing fact (even if in a less general context).

Question: Do I really need $T$ being kernel preserving? Is there a way to deduce it from the hypotheses?

Ender Wiggins
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  • It's not necessarily true that $T$ preserves kernels: we could have $T=T_0\oplus T_1$ where $T_0$ is faithful and $T_1$ does not preserve kernels. That will still reflect exact sequences if $T_0$ does, though. – Eric Wofsey Aug 01 '19 at 17:08
  • @EricWofsey So, are you saying that I should be able to prove that a functor is faithful if and only if it reflects exact sequences without additional requirements? – Ender Wiggins Aug 02 '19 at 05:51
  • Well, assuming the exercise is correct. I haven't been able to come up with a proof myself. The difficulty is that images are defined by a compound of two universal properties (kernel of a cokernel or cokernel of a kernel), so it seems hard to usefully relate the image of $f$ and the image of $T(f)$. – Eric Wofsey Aug 02 '19 at 05:55
  • I had the same problem, but with kernels. By faithfulness of $T$ exists $\lambda:\mathrm{im}(f)\to \ker(g)$ and I want to prove that it is bijective. If $(K,k)$ is the kernel of $g$, $(I,i)$ the image of $f$ and $(J,j)$ is $\mathrm{im}(Tf)=\ker(Tg)$ then there exist unique $\tau: T(K) \to J$, $\mu: T(I)\to J$ such that $j\mu = T(i)$ and $j\tau = T(k)$. Using that $j$ is monic one can conclude that $\tau T(i) = \mu$ but then without knowing at least that $T(\mathrm{im}(f))$ is the kernel of $Tg$ I don't know how to proceed. – Ender Wiggins Aug 02 '19 at 06:07
  • Yeah, that's pretty much the same as what I was trying. My idea was to then get a map from $J$ to $T(I)$ using the fact that $J$ is the cokernel of the kernel of $T(f)$ (the cokernel property gives us a way to get maps out of it), but couldn't get anywhere because $T$ might not preserve the kernel of $f$ (indeed, the map $T(A)\to T(I)$ may not even annihilate the kernel of $T(f)$, in an example as in my first comment). – Eric Wofsey Aug 02 '19 at 06:19
  • That's what I meant by the difficulty coming from the compound universal property: I can't use the cokernel universal property of $J$ because it's not just a cokernel but a cokernel of a kernel, and I don't have enough control first over the kernel. – Eric Wofsey Aug 02 '19 at 06:22

1 Answers1

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Here is a proof. Suppose $$T(A) \stackrel{T(f)}{\longrightarrow} T(B)\stackrel{T(g)}{\longrightarrow} T(C)$$ is exact. Then $T(gf)=0$ so $gf=0$.

Now let $k: K\to B$ be a kernel of $g$, let $j:J\to T(B)$ be a kernel of $T(g)$, let $d:B\to D$ be a cokernel of $f$, and let $e:T(B)\to E$ be a cokernel of $T(f)$. Since $T(A)\to T(B)\to T(C)$ is exact, $j$ is also the inclusion of the image of $T(f)$ and thus $e$ is also a cokernel of $j$; in particular, $ej=0$. Since $T(g)T(k)=0$, there is $\alpha:T(K)\to J$ such that $j\alpha=T(k)$. Similarly, there is $\beta:E\to T(D)$ such that $\beta e=T(d)$. Thus $T(dk)=\beta e j\alpha=0$. Since $T$ is faithful, $dk=0$. But that says exactly that the image of $f$ is contains the kernel of $g$, and so $A\to B\to C$ is exact.

Note that these hypotheses do not imply that $T$ preserves kernels. For instance, $T:Ab\to Ab$ could be the functor $T(A)=A\oplus A/2A$, which is faithful because of the first summand but does not preserve kernels because of the second. More generally, $T$ could have the form $T_0\oplus T_1$ where $T_0$ is faithful and $T_1$ is an arbitrarily ill-behaved additive functor.

Eric Wofsey
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