Existence of adjugant with making given natural transformation be the counit

55 Views Asked by At

I’m studying a category theory with the book Hilton-Stammbach’s “A course in homological algebra”. On solving the exercise problems in the adjoint functor section(Chapter 2, section 7), I had had a hard time understanding what the following problem wants me to prove.

Exercise 7.5(p.69) Show that it is possible to choose, for each $R$-module $M$, a surjection $P(M) \xrightarrow{\epsilon_M}$ M, where $P(M)$ is a free $R$-module, in such a way that $P$ is a functor from a category of $R$-modules Mod$_R$ to a category of free $R$-modules FMod$_R$ and $\epsilon_M$ is a natural transformation. If $E:$FMod$_R$$\to$ Mod$_R$ is the embedding functor, is there an adjugant $\eta: E \dashv P$ such that $\epsilon$ is the counit?

Here, $R$ is a ring with $1_R \neq 0$. For the first part of the problem, I guessed the following statement,

Every $R-$module is a quotient of a free $R$-module.

may be helpful, but no such a progression was made. Since I stuck at the very starting point of this problem, it will be really appreciated if you give a small hint or nudge for solving this problem. Thank you.

1

There are 1 best solutions below

0
On BEST ANSWER

Hint: You want to use not just the fact that every $R$-module is a quotient of a free $R$-module, but the proof of this fact. In particular, can you describe a "canonical" way to make any $R$-module a quotient of a free $R$-module? That is, if I hand you a module $M$, how could you construct a free module which $M$ is a quotient of, without making any arbitrary choices?

Once you have that, then you can try to turn this construction into a functor. That is, you want to define a functor $P$ which on objects takes a module to the "canonical" free $R$-module it is a quotient of. To do this you'll have to come up with a natural thing that $P$ could do on arrows, and prove that it works. Then, you want to check that if you let $\epsilon_M$ be the quotient map from $P(M)$ to $M$, this really is a natural transformation.

A stronger hint (answering the question of the first paragraph above) is hidden below.

Take $P(M)$ to be the free module on the underlying set of $M$.