Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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.

share|improve this answer
    
Thank you, that was very helpful. –  PavelC Jun 18 '13 at 15:49

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.