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While defining Adjoint functors in their book A Course in Homological Algebra, Hilton and Stammbach said the follwing:

Let $F:\mathfrak{S}\rightarrow \mathfrak{M}_{\Lambda}$ be the free functor which associates with every set the free $\Lambda$ module on that set as a basis; and let $G:\mathfrak{M}_{\Lambda}\rightarrow\mathfrak{S}$ be the underlying functor which associates with every module its underlying set. We now define a transformation, natural in both $S$ and $A$, $$\eta=\eta_{SA}:\mathfrak{M}_{\Lambda}(FS,A)\rightarrow \mathfrak{S}(S,GA)$$

I did not understand what does he mean by natural in both $S$ and $A$..

I know that a natural transformation is between two functors $F,G: \mathcal{C}\rightarrow \mathcal{D}$..

What are those two functors here...

Fix $A$ a $\Lambda$ module. Then we have functor $$\mathfrak{M}_{\Lambda}(F-,A):S\mapsto \mathfrak{M}_{\Lambda}(FS,A)$$ from sets to module homomorphisms

Fix $S$ a set. Then we have a functor $$\mathfrak{S}(S,G-):A\mapsto \mathfrak{S}(S,GA)$$ from modules to set maps..

To define natural transformations, we need to have two functors from same categor landing in same category... Here one functor is from category of sets and other is from category of modules... What natural equivalence are we considering here?

I am very much confused..

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There seems to be a misunderstanding here. Your $\mathfrak M_Λ(F-, A)$ is not a functor from sets to "module homomorphisms", it is a contravariant functor from sets to sets, which assigns to every set $S$ a set of certain module homomorphisms. In other words, it is a functor from $\mathfrak S^\mathrm{op}$ to $\mathfrak S$, as is $\mathfrak S(-, GA)$. That $η_{SA}$ is natural in $S$ means that for a fixed $A$, $η_{-,A}$ is a natural transformation between these two functors, and similarly for naturality in $A$.

Another way to say this is to note that $(S, A) ↦ \mathfrak M_Λ(FS, A)$ is a functor $\mathfrak S^\mathrm{op} × \mathfrak M_Λ → \mathfrak S$, and so is $(S, A) ↦ \mathfrak S(S, GA)$, and you can check that $η$ is natural in both $S$ and $A$ exactly when it is a natural transformation between them.

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  • $\begingroup$ I think i got it... Thanks.... Give me some time to understand it better... If i have no more questions i will upvote it.. $\endgroup$ – kk lm Feb 3 '16 at 4:35
  • $\begingroup$ I am still getting confused... Why do we have $\mathfrak{S}^{\rm{op}}$ there and not just $\mathfrak{S}$ $\endgroup$ – kk lm Feb 3 '16 at 19:04
  • $\begingroup$ Because the hom functor is contravariant in the first argument, which is surely explained in a textbook on homological algebra; try revisiting the previous parts on categories. $\endgroup$ – user54748 Feb 3 '16 at 19:16
  • $\begingroup$ Yes Yes.. Hom functor is contravariant in the first argument... I will revise what i have learned before and then get back to this.. Thanks... $\endgroup$ – kk lm Feb 3 '16 at 19:19

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