Show that if $$0 \rightarrow M_1 \xrightarrow{f} M_2 \xrightarrow{g} M_3$$ is an exact sequence of $R$-modules, then for all $R$-module $$0 \rightarrow \operatorname{Hom}_R(M,M_1) \xrightarrow{f_*} \operatorname{Hom}_R(M,M_2) \xrightarrow{g_*} \operatorname{Hom}_R(M,M_3)$$is an exact sequence of $\mathbb Z$-modules ($f_*(h)=f \circ h$).

In order to show this is an exact sequence, I have to prove $Im(f_*)=Ker(g_*)$. I could show the inclusion $Im(f_*) \subset Ker(g_*)$.Let $\psi \in Im(f_*)$, then there is $h \in Hom_R(M,M_1) : f \circ h=\psi$. It is clear that $\psi \in Hom_R(M,M_2)$. Let's evaluate in $g_*$:$g_*(\psi)=g \circ \psi=g \circ f \circ h$. We want to show $\psi \in Ker(g_*)$, i.e., $g \circ \psi \equiv 0$. Let $x \in M$, then $g \circ \psi (x)= g \circ f\circ h(x)=g(f(h(x)))$. But $Im(f)=Ker(g)$, so $g(f(h(x))))=0$. From here one can conclude $\psi \in Ker(g_*)$.

I couldn't show the other inclusion, any help would be appreciated.

  • 1
    $\begingroup$ Note that you also have to show that $f_*$ is injective. $\endgroup$
    – jflipp
    Dec 11, 2014 at 14:01
  • $\begingroup$ Just to check that: take $h,g \in Hom_R(M,M_1)$ and suppose $f \circ h= f \circ g$, then $f(h(x)-g(x))=0$ for $x$ arbitrary, but then $h(x)-g(x) \in Ker(f)=\{0\}$, so $h(x)=g(x)$ for all $x \in M$, it follows $f_*$ is injective. $\endgroup$
    – user16924
    Dec 11, 2014 at 19:24

1 Answer 1


We can argue as follows. Pick an $h \in ker\ g_* \subseteq Hom_R(M,M_2).$ Then we have $g\circ h = 0.$ This means $im\ h \subseteq ker\ g = im\ f.$ Since $f$ is injective, we have the $R$-homomorphism $(f|_{im\ f})^{-1}:im\ f \rightarrow M_1.$ So we can define $j := (f|_{im\ f})^{-1} \circ h: M \rightarrow M_1.$ This is an $R$-homomorphism since it's a composition of $R$-homomorphisms. By construction, we have $f\circ j = h,$ or, in other words $h = f_*(j).$ Since $h$ was chosen arbitrarily, we have shown $ker\ g_* \subseteq im\ f_*,$ as desired.


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