# Is $D=\hom_K(\cdot, K)$ exact?

Let $K$ be an algebraically closed field. On page 31 of the book Elements of representation theory of associative algebras, volume 1, from Theorem 5.13 (a) we see that $D=\hom_K(\cdot, K)$ is exact. But we know that $\hom$ is left exact and not necessarily exact. I am confused about this. Is it because $D=\hom_K(\cdot, K)$ but not $\hom_A(\cdot, A)$, where $A$ is a $K$-algebra. Thank you very much.

• $K$ is obviously injective as $K$-module. Actually $\mathrm{Hom}_K(-,V)$ is exact for any $K$-module $V$, because $K$ is a field, so a semisimple ring. – egreg May 23 '13 at 12:56
• @egreg Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. – Julian Kuelshammer Jun 19 '13 at 6:01
• @JulianKuelshammer Done. – egreg Jun 19 '13 at 8:38

Since $K$ is a field, it's in particular semisimple and every $K$-module is injective, which amounts to say that $$\operatorname{Hom}_K(-,V)$$ is exact for any $K$-module $V$.
A doubt could arise because the functor $D=\operatorname{Hom}_K(-,K)$ is considered as $D\colon\operatorname{mod} A\to\operatorname{mod}A^{\mathrm{op}}$, but this is irrelevant; what is needed is only that, when $M\to N$ is monic in $\operatorname{mod}A$, the morphism $$\operatorname{Hom}_K(N,K)\to\operatorname{Hom}_K(M,K)$$ is epic, which is true.