# Is the Auslander-Reiten translation, restricted to non-projective and non-injective modules, exact?

I guess there should be a general category-theory answer to this question, but I'm asking what I'm interested in specifically, since I don't know if it is true:

Let $H$ be a finite-dimensional hereditary algebra. Let $\tau = D \circ Tr: \text{mod}(H) \to \text{mod}(H)$ be the Auslander-Reiten translation and let $\tau^{-} = Tr \circ D: \text{mod}(H) \to \text{mod}(H)$ be the Auslander-Reiten translation in the other direction.

Fact 1: $\tau$ is a left exact additive functor, $\tau^-$ is a right exact additive functor.

Fact 2: Let $\text{mod}(H)_i$ be the category of finite-dimensional $H$-modules without indecomposable injective direct summands and $\text{mod}(H)_p$ be the category of finite-dimensional $H$-modules without indecomposable projective direct summands. Then $\tau: \text{mod}(H)_p \to \text{mod}(H)_i$ is an equivalence of categories with quasi-inverse $\tau^{-}$.

Question: Are $\tau$ and $\tau^-$, restricted as above, exact?

Added after Jeremy's answer: Is there also a proof which only uses the two facts stated by me and not the concrete definition of the Auslander-Reiten translation? So is this somehow just a category theory consequence?

• Your title mentions (non-)preprojective and preinjective modules, but they don't seem to play a role in the question? – Jeremy Rickard Sep 20 '17 at 8:58
• I'm sorry. Originally I thought it would be about non-preprojective (and non-preinjective) modules and while formulating the question I realized it is about non-projective and non-injective modules. I changed the title now. – Leon Lang Sep 20 '17 at 18:24

If $H$ is hereditary, then Auslander-Reiten translation is simply $\tau X=D\text{Ext}^1_H(X,H)$. Also, $\text{Ext}^2_H$ vanishes, and if $X$ has no nonzero projective direct summand then $\text{Hom}_H(X,H)=0$ since every submodule of $H$ is projective. [If $\alpha:X\to H$ is a non-zero map then $\text{im}(\alpha)$ is projective, so $X\to\text{im}(\alpha)$ splits, and so $X$ has a direct summand isomorphic to $\text{im}(\alpha)$.]
So if $0\to L\to M\to N\to 0$ is a short exact sequence of $H$-modules with no projective summands, the long exact $\text{Ext}$ sequence obtained by applying $\text{Hom}_H(-,H)$ reduces to $$0\to\text{Ext}^1_H(N,H)\to\text{Ext}^1_H(M,H)\to\text{Ext}^1_H(L,H)\to0$$ and so, taking vector space duals, we get a short exact sequence $$0\to\tau L\to\tau M\to\tau N\to0.$$