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Let $E^\bullet$ and $F^\bullet$ be complexes on an abelian category; what does it mean to say that $E^\bullet$ and $F^\bullet$ are quasi-isomorphic?

Does it only mean that there is a map of complexes $f:E^\bullet \to F^\bullet$ that induces isomoprhisms between the cohomology objects?

Or does it also guarantee the existence of a map of complexes $g:F^\bullet \to E^\bullet$ inducing the inverses of $H^pf:H^p(E^\bullet)\to H^p(F^\bullet)$?

Put in another way: is quasi-isomorphism an equivalence relation?

t.b.
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Marcos Jardim
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  • I tried to edit, but apparently I have to change 6 characters. The map should say $g:F^\bullet \to E^\bullet$. – Matt Dec 21 '11 at 19:26
  • You could be interested in http://en.wikipedia.org/wiki/Derived_category – Alexander Thumm Dec 21 '11 at 19:39
  • Does the equivalence relation "two complexes E and F are equivalent if there are quasi-isomorphisms f:E-->F and g:F-->E" have a particular name in the literature? – Marcos Jardim Dec 22 '11 at 14:01

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$\def\ZZ{\mathbb Z}$The relation $E \sim F$ defined by «there exists a morphism $E\to F$ inducing an isomorphism in homology» is not an equivalence relation because it is not symmetric (it is relfexive and transitive)

For example, there is a morphism from $$\cdots 0\to \ZZ\xrightarrow2\ZZ\to0\to\cdots$$ to the complex $$\cdots 0\to 0\to\ZZ/2\ZZ\to0\to\cdots$$ inducing an isomorphism in homology, but there is no non-zero morphism in the other direction.

The useful relation is the symmetric closure of this relation.

Mariano Suárez-Álvarez
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  • Thank you for the nice counter-example. Does the equivalence relation "two complexes E and F are equivalent if there are quasi-isomorphisms f:E-->F and g:F-->E" have a particular name in the literature? – Marcos Jardim Dec 22 '11 at 14:00