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Let $\mathcal A$ be an abelian category. Let $\mathbb K(\mathcal A)$ be the homotopy category of complexes of $\mathcal A$. Now. consider a functor $F$ from $\mathbb K(\mathcal A)$ to $\mathbb E$, which is also a triangulated category. Suppose the $F$ takes quasi-isomorphisms to isomorphisms. Now by the universal property of $\mathbb D(\mathcal A)$, the derived category of $\mathcal A$, we know there is a functor $F'$ from $\mathbb D(\mathcal A)$ to $\mathbb E$ induced by functor $F$.

Is $F'$ a triangulated functor?

By triangulated functor, I mean a functor which commutes with the translation functor and takes distinguished triangles to distinguished triangles.

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If you don't assume that $F$ is a triangulated functor then this may not be true.

For example, let $F:\mathbb{K}(\mathcal{A})\to\mathbb{K}(\mathcal{A})$ be the obvious functor sending a complex to its degree zero homology considered as a complex concentrated in degree zero. Then $F'$ doesn't commute with the translation functor.

But if you intended to specify that $F$ is triangulated, then it is true, as explained in Pierre-Guy Plamondon's answer.

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Edit: In this answer, I assume that the functor $F$ is triangulated. The answer is different in case $F$ is not, as explained in Jeremy Rickard's answer.

Yes, $F'$ is a triangulated functor.

One way to formulate this is by saying that the derived category is the triangulated quotient of $\mathbb{K}(\mathcal{A})$ by the full subcategory of objects with vanishing total homology. The universal property of a triangulated quotient then gives the property you want, since any functor $F$ that takes quasi-isomorphisms to true isomorphisms will take objects with vanishing homology to zero.

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    $\begingroup$ You're assuming that $F$ is triangulated, which the OP didn't specify. But there seems a good chance that he meant to, so I guess our two answers cover all bases! $\endgroup$ – Jeremy Rickard Aug 11 '16 at 16:04
  • $\begingroup$ @JeremyRickard You're right, the OP did not assume it, and I did... But with your answer, the gap is filled, so indeed we should be good! $\endgroup$ – Pierre-Guy Plamondon Aug 11 '16 at 20:22

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