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I'm trying to find an exact sequence such that $Hom (\mathbb{\prod Z}, -)$ is not exact, I tried to put $ L(U (\prod \mathbb{Z})) \twoheadrightarrow \prod \mathbb{Z}$ where $L$ is left adjoint to $U$ and $U$ forgets the group structure, but apparently it does work for the identity arrow, so it does not show that the projectiveness fails. Any hint or answer?

Thanks in advance.

user40276
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  • I am baffled by your comment, that it "works" for the identity arrow. If it did, that would furnish a splitting of the counit which would immediately imply projectivity of $\prod \mathbb{Z}$ (I take it you mean an infinite product here). Meanwhile, we know for other reasons that the infinite product is not projective -- is your real question why that is so? – user43208 Oct 04 '13 at 17:46
  • @user43208 I know it's not projective (it follows from the fact that it's not free), however I was trying to explicitly show that it's not projective. – user40276 Oct 04 '13 at 17:52
  • Is it not clear that there can be no element $s \in \hom(\prod \mathbb{Z}, LU \prod \mathbb{Z})$ that maps to the identity under the induced map $\hom(\text{id}, \pi): \hom(\prod \mathbb{Z}, LU \prod \mathbb{Z}) \to \hom(\prod \mathbb{Z}, \prod \mathbb{Z})$? Because if there were, then that would say $\pi \circ s = \text{id}$, meaning that $s$ gives a splitting that would exhibit $\prod \mathbb{Z}$ as a direct summand of the free module $LU \prod \mathbb{Z}$. – user43208 Oct 04 '13 at 17:59

1 Answers1

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It is well-known that over a PID such as $\mathbb{Z}$, a submodule of a free module is free. In particular, projective modules are the same as free modules in this case.

As in the OP, let $L: \text{Set} \to \text{Mod}_\mathbb{Z}$ be the free functor which is left adjoint to the forgetful functor $U: \text{Mod}_\mathbb{Z} \to \text{Set}$. There is a canonical counit map $\pi_A: LU(A) \to A$ for any module $A$; it sends each generator $a$ of $LU(A)$ (coming from the underlying set of $A$) to $A$. It is thus surjective.

If $\hom(A, -)$ preserves the exact sequence $LU(A) \stackrel{\pi_A}{\to} A \to 0$, then the induced map

$$\hom(A, LU(A)) \stackrel{\hom(1_A, \pi)}{\to} \hom(A, A)$$

would be surjective, so that there would exist $s \in \hom(A, LU(A))$ mapping onto the identity $1_A: A \to A$. By definition of the hom-functor $\hom(A, -)$, this means precisely that $\pi_A \circ s = 1_A$. This forces $s$ to be a monomorphism (since $1_A$ is of course a monomorphism), so that $s: A \to LU(A)$ maps $A$ isomorphically onto its image, which is a subgroup of the free abelian group $LU(A)$, making $A$ itself free abelian.

In particular, if we know that $A = \prod \mathbb{Z}$ (an infinite power of $\mathbb{Z}$) is not free abelian (which in itself is an interesting result), then by the above we have exhibited an exact sequence that is not preserved by $\hom(A, -)$.

user43208
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  • I just got confused when I thought about the map $\varphi (1.x) = x$ where $1.x$ means the value $x$ in the coordinate $1$ (as a basis), but then I realized that it's not a homomorphism. – user40276 Oct 19 '13 at 22:53