Library of Mathexandria

From the previous post, we have considered $Hom_R(B, -)$, but this time we want to consider $Hom_R(-,A)$. Instead of getting a covariant functor, we now get a contravariant functor (i.e. reverses the map). Consider a map $f: P_1 \rightarrow P_2$, then the natural way to pass it to Hom-sets are as follows, $f^*: Hom_R(P_2, A) \rightarrow Hom_R(P_1, A) (\dagger)$ $\phi \mapsto \phi \circ f$ If we try to construct the order way,  $Hom_R(P_1, A) \rightarrow Hom_R(P_2, A)$ $\phi \mapsto \phi \circ f$? $\phi \mapsto f \circ \phi$? Both composition would not make sense, hence we take $(\dagger)$ as our $Hom_R(-,A)$ functor. And obviously $R$-module $A$, $P$ is sent to $Hom_R(P,A)$. We now prove that $Hom_R(-,A)$ functor is again a left-exact functor, which means the following, Given a short exact sequence, $0 \rightarrow P_1 \stackrel{f}{\rightarrow} P_2 \stackrel{g}{\rightarrow} P_3$ $0 \rightarrow Hom_R(P_3,A) \stackrel{g^*}{\rightarrow} Hom_R(P_2, A) \stackrel{f^*}{\rightarrow} Hom_R(...

In this post, we prove the well-used property that the Hom functor is left-exact. We consider the Hom functor of the following form $Hom_R(B,-)$ which is a functor from $R-mod$ the category of $R$-modules to $Ab$ the category of abelian groups. The functor contains the following data: $A \mapsto Hom_R(B,A)$ $\{ f: A \rightarrow A' \} \mapsto \{ f^*: Hom_R(B,A) \rightarrow Hom_R(B,A') \}$ where $f^*$ is defined by $f^*(\phi) = f \circ \phi$. Then we see that Hom functor is a covariant functor.   Proof) If $f : A \rightarrow A$ is the identity map, then $f^*(\phi) = f \circ \phi = f$, hence $f^*$ also becomes an identity map of $Hom_R(B,A)$. Also, let $f : A \rightarrow A'$ and $g: A' \rightarrow A''$, then we see that  $(g \circ f)^*(\phi) = (g \circ f) \circ \phi  = g \circ (f \circ \phi) = g \circ (f^*(\phi)) = (g^* \circ f^*)(\phi)$.  i.e. $Hom_R(-,B)$ is a covariant functor. Now to our main point that $Hom_R(B,-)$ is left-exact. A functor $\mathcal{F}$ is le...