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Suppose I have an $R$-module $M$ and a free resolution $$ \ldots \to F_2 \to F_1 \to M \to 0. $$ I apply an additive functor $f$ in $R$-$\mathbf{Mod}$ to the free resolution to get $$ \ldots \to f(F_2) \to f(F_1) \to f(M) \to 0. $$ This is a chain complex, so I can take the homology of this. Will the homology be independent of the free resolution that I started with?

I believe the answer is yes. Here's my rough reasoning. First of all, the sequence $f(F_\bullet)$ is a chain complex because $f$ brings zero morphisms to zero morphisms by additivity. Any chain map between two free resolutions remains a chain map under $f$. Also, any chain homotopy between two chain maps remains a chain homotopy under $f$.

However, in all sources I have found, they would require $f$ to be right-exact, and the resulting homology after applying $f$ would be referred to as the left derived functor $L^i$. I am not sure if right-exactness is necessary in making the homology well-defined, or if it is needed to show that each $L^i$ is a functor, or if it is only needed to show that the sequence of $L^i$ forms a $\delta$-functor.

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Exactness is needed for universality, I think. – Zhen Lin Jan 17 '14 at 13:50
You are correct that the homology is independent of the resolution. The main difference is that you will not have $H_0(F_\bullet) = f(M)$ unless $f$ is exact, but it will instead be something called the zero'th left derived functor of $f$. Some cases where this has shown up in real-world applications include where $f$ is completion with respect to an ideal. However, in some cases (e.g. where $f$ is left exact) you might not get an interesting answer. – Tyler Lawson Jan 17 '14 at 14:14
up vote 4 down vote accepted

I think this quote answers your question:

''Theorem $5.3$ shows that the right derived functors are of real interest only if $T$ is left exact. Indeed $R^n T$ may always be replaced by $R^n T'$ with $T' = RT$ which is left exact. Similarly the left derived functors are mainly interesting for functors which are right exact.''

[Cartan E., Eilenberg S. Homological Algebra, p.90]

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Thank you for reminding me of the existence of this wonderful source. How ignorant was I to have overlooked it! – Tunococ Jan 17 '14 at 22:35
I guess I'm older than you, so I taught Homological Algebra just by this book :-) – Boris Novikov Jan 18 '14 at 7:54

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