Trying to understand some formulas about dualities, specifically $\operatorname{Hom}_{\Lambda}(Y,\nu X) \cong D\operatorname{Hom}_{\Lambda}(X,Y)$.

Question

I'm reading $\tau$-tilting by Adachi, Iyama and Reiten. There is a particular isomorphism of hom-sets which I'm having trouble understanding.

The setup:

$\Lambda$ is a finite-dimensional $k$-algebra, where $k$ is an algebraically closed field. Let $X,Y$ be finitely generated $\Lambda$-modules. We have the nakayama functor $\nu = D\operatorname{Hom}_{\Lambda}(-,\Lambda)$. Then, they write $$\operatorname{Hom}_{\Lambda}(Y,\nu X) \cong D\operatorname{Hom}_{\Lambda}(X,Y)$$ which I just don't get.

Here's my attempt at trying to understand what's going on. I get lost pretty quickly in the details.

Regarding $\operatorname{Hom}_{\Lambda}(Y,\nu X)$:

Let $f \in \operatorname{Hom}_{\Lambda}(X,\Lambda)$ and apply $D = \operatorname{Hom}_{k}(-,k)$ to it. Then we have $$\operatorname{Hom}_{k}(\Lambda, k) \xrightarrow{f^{\ast}} \operatorname{Hom}_{k}(X, k)$$ by $f^{\ast}(\phi) = \phi f$ for any $\phi \in \operatorname{Hom}_{k}(\Lambda, k)$. Now, $f^{\ast} \in \nu X$.

Let $g \in \operatorname{Hom}_{\Lambda}(Y, \nu X)$ such that $g(y) = f^{\ast}$, so for any $\phi \in \operatorname{Hom}_{k}(\Lambda, k)$ we have $g(y)(\phi) = f^{\ast}(\phi) \in \operatorname{Hom}_{k}(X,k)$

Regarding $D\operatorname{Hom}_{\Lambda}(X,Y)$:

Let $h \in \operatorname{Hom}_{\Lambda}(X,Y)$ and apply $D = \operatorname{Hom}_{k}(-,k)$ to it. Then we have $$\operatorname{Hom}_{k}(Y,k) \xrightarrow{h^{\ast}} \operatorname{Hom}_{k}(X,k)$$ by $h^{\ast}(\psi) = \psi h$ for any $\psi \in \operatorname{Hom}_{k}(Y,k)$. Now $h^{\ast} \in D\operatorname{Hom}_{\Lambda}(X,Y)$.

I can't see what the isomorphism is, i.e. how to send some $g$ to some $h^{\ast}$ and vice-versa as defined above. Could someone shed some light on this?

Answer 1

Let A be the quiver algebra with two points 1 and 2 and one arrow from 1 to 2. The simple module $S_1$ has $\nu (S_1)=0$. Take $X=Y=S_1$, so $DHom_A(S_1,S_1)$ is one-dimensional but $Hom_A(S_1 , \nu(S_1))$ is zero.

Where exactly in the article did you see that? You probably miss an assumption. There is indeed a natural isomorphism in case $X$ or $Y$ is projective, see for example the book Frobenius algebras I by Skowronski and Yamagata in Lemma 6.1. of Chapter III.