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A morphism of triangulated categories is often called "exact" if it preserves the shift operator and takes distinguished triangles to distinguished triangles.

Are exact functors between triangulated categories necessarily exact as functors? (Ie, do they preserve finite limits and colimits?)

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    $\begingroup$ My initial feeling is that this is an ill-posed question because triangulated categories don't have well-behaved limits and colimits. What's your motivation for the question? $\endgroup$
    – Hanno
    Jul 8, 2020 at 10:56

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Yes they are, but in fact any additive functor between triangulated categories preserve finite limits and colimits. I will only give the main ingredients for the proof, let me know if you want me to expand on certain points. So let $F$ be any additive functor.

An additive functor preserves finite limits iff it preserves kernels and cokernels

This is because it preserves finite finite direct sum and finite direct products, and because in additive categories preserving kernels is equivalent to preserving equalizer, and preserving cokernels is equivalent to preserving coequalizers. Let us show that $F$ preserves kernels, by duality it will preserve cokernels.

In a triangulated categories, monomorphisms splits.

It follows that if $f:A\to B$ has a kernel, then you have an object $I$ such that $A\simeq \ker f\oplus I$. Let us denote $k:\ker f\to A$ and $i:I\to A$ the canonical inclusions and $p:A\to \ker f$, $q:A\to I$ the canonical projections.

The morphism $fi:I\to B$ is a monomorphism.

Indeed, if $X\xrightarrow{u} I\to B$ is zero, then $X\xrightarrow{u} I\xrightarrow{i} A\xrightarrow{f} B$ is zero and $X\xrightarrow{u} I\xrightarrow{i} A$ factors through $k$. So we have $iu=ku'$. Hence $piu=pku'$ which is $0=u'$ (since $pi=0$ and $pk=id$). You thus have $iu=0$ and since $i$ is a monomorphism (it splits), $u=0$. As above, we can write $B\simeq I\oplus C$ (the inclusion $I\to B$ being $fi$).

We now have $f=fiq$ is a composite of a split mono ($fi$) and a split epi $q$.

Any functor preserve split monos and split epis

Hence we still have $F(A)=F(\ker f)\oplus F(I)$ and $F(B)=F(I)\oplus F(C)$ and $F(f)$ is the composite $F(A)\to F(I)\to F(B)$. In this situation $F(f)$ has a kernel and is canonically $F(\ker f)$. So the canonical map $\ker F(f)\to F(\ker f)$ is an isomorphism.


As a side note, in some text distinguished triangles are called exact triangles. A functor between triangulated categories is thus called exact if it preserves exact triangles. But a better terminology would be to called this triangles homotopically exact, and thus speaking about homotopically exact functors.

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