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I am trying to to understand the definition of derived functors in the context of derived categories, primarily using Huybrechts' book Fourier-Mukai transforms in algebraic geometry, but I have consulted other sources as well. I have some experience with Ext and Tor functors, so this is not entirely new to me.

Let $F:\mathcal{A} \to \mathcal{B}$ be a left exact additive functor between abelian categories, where $\mathcal{A}$ has enough injectives. This allows us to define the derived functor $RF:D^+(\mathcal{A}) \to D^+(\mathcal{B})$, where $D^+(\mathcal{A})$ is the bounded-below derived category. Then for $n \ge 0$, Huybrechts defines $$ R^n F:D^+(\mathcal{A}) \to \mathcal{B} \qquad R^nF(X^\bullet):=H^n(RF(X^\bullet)) $$ where $X^\bullet$ is a chain complex over $\mathcal{A}$ and $H^n$ is the homology functor. I'm fine with everything up to this point. Then, Huybrechts says, "The induced additive functors $R^nF:\mathcal{A} \to \mathcal{B}$ are the higher derived functors of $F$."

How do we get an induced functor $\mathcal{A} \to \mathcal{B}$?

I know that the right answer "should" be that for an object $X$ of $\mathcal{A}$, we choose an injective resolution of $X$, chop off $X$, view it as a bounded below chain complex (object of $D^+(\mathcal{A})$), then apply $R^n F$ as defined on $D^+(\mathcal{A})$. It is clear from the discussion following the definition that this is also what Huybrechts has in mind.

However, I also know from past experience that showing that such a definition is independent of the choice of injective resolution is quite technical and tedious. Since Huybrechts doesn't even mention this possible ambiguity, perhaps I'm misunderstanding some of the previous lemmas, and that they somehow imply that this isn't a concern.

Can we define the induced functor $\mathcal{A} \to \mathcal{B}$ in such a way that this ambiguity is resolved without the details of the "Fundamental Lemma of Homological Algebra"?

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There is a functor $i:A\to D^+(A)$ composed from the embedding of $A$ into chain complexes. Huybrechts' functors just come from composing $R^nF$ with $i$. There's nothing more to the definition than that-in particular, there's no need to mention injective resolutions at this point.

The work of showing that this is well defined is shifted, relative to the classical viewpoint, into the work of showing that $D^+A$ and $RF$ exist; after that, it's a triviality that any two injective resolutions of an object of $A$ are isomorphic in $D^+A$, and every functor preserves isomorphisms.

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  • $\begingroup$ By embedding $\mathcal{A} \to D^+(\mathcal{A})$, you mean we view an object $X$ of $\mathcal{A}$ as a chain complex supported in dimension zero? $\endgroup$ – Joshua Ruiter Mar 23 '18 at 1:55
  • $\begingroup$ Yes, that's right. $\endgroup$ – Kevin Carlson Mar 23 '18 at 14:36

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