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For simplicity's sake, consider the categories $R\text{-Mod}, S\text{-Mod}$ of left $R$-modules and left $S$-modules, respectively, and let $\mathcal{F}$ be some precovering class in $R\text{-Mod}$. Then given a functor $T:R\text{-Mod} \rightarrow S\text{-Mod}$, left derived functors $L_nT$ can be defined.

In a book I recently read (Relative homological Algebra by Enochs and Jenda), a canonical natural transformation $L_0T\rightarrow T$ is mentioned but not described.

When I tried to construct such transformation, it occured to me that there is a natural transformation going in the opposite direction:

Given a left $R-$module $M$, it is easy to see that $L_0T(M)$ is simply a factor of $T(M)$ modulo $ImT(f)$ for some morphism $f:F\rightarrow M$ (the "beginning" of left $\mathcal{F}$-resolution of $M$). Then it is easy to check that $\tau_M:T(M) \rightarrow L_0T(M)$ defined via $\tau_M(x)=x+ImT(f)$ is a natural transformation.

So my question is:

Is it possible, that this is the intended natural transformation and the direction was just reversed by mistake? If not, is there a general way of defining a natural transformation $\sigma: L_0T\rightarrow T$, such that, assuming $T$ is right exact, it is an isomorphism?

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If you write R-Mod in TeX within math mode, it comes out as $R-Mod$ with the hyphen looking like a minus sign instead of a hyphen. I changed it to $R\text{-Mod}$, coded as R\text{-Mod}. –  Michael Hardy Jun 17 '13 at 18:17

1 Answer 1

up vote 1 down vote accepted

It seems to me that this is the right direction. To make sure we are on the same page let me recall how I remember the relevant definitions.

In the context of module categories, one can define the left derived functors as $L^{i}F(M) = H^{i}(F(P^{\bullet}))$, where $P^{\bullet}$ is a projective resolution of $M$. We say $P^{\bullet}$ is a projective resolution if it is a complex vanishing in negative degrees such that all its objects are projective, together with a map $q: P^{\bullet} \rightarrow M$ that induces isomorphisms in homology. (Where we see $M$ as a complex concentrated in degree $0$).

Hence, the induced map $H^{0}(T(q)): L^{0}F(M) \simeq H^{0}(T(P^{\bullet})) \rightarrow H^{0}(T(M)) \simeq T(M)$ seems to be the required natural transformation.

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Thank you, that was very helpful. –  PavelC Jun 18 '13 at 15:49

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