# Showing that $0 \to \mathrm{Hom}(M, M_1) \to \mathrm{Hom}_R(M, M_2) \to \mathrm{Hom}(M, M_3)$ is exact

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 $$M$$, $$0 \rightarrow \operatorname{Hom}_R(M,M_1) \xrightarrow{f_*} \operatorname{Hom}_R(M,M_2) \xrightarrow{g_*} \operatorname{Hom}_R(M,M_3)$$ is an exact sequence of $$\mathbb{Z}$$-modules (where $$f_*(h)=f \circ h$$).

In order to show this is an exact sequence, I have to prove $$\mathrm{Im}(f_*) = \mathrm{Ker}(g_*)$$.

I could show the inclusion $$\mathrm{Im}(f_*) \subset \mathrm{Ker}(g_*)$$: Let $$\psi \in \mathrm{Im}(f_*)$$, then there is $$h \in \mathrm{Hom}_R(M, M_1)$$ with $$f \circ h = \psi$$. It is clear that $$\psi \in \mathrm{Hom}_R(M, M_2)$$. Let’s evaluate in $$g_*$$: $$g_*(\psi) = g \circ \psi = g \circ f \circ h \,.$$ We want to show $$\psi \in \mathrm{Ker}(g_*)$$, 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 $$\mathrm{Im}(f) = \mathrm{Ker}(g)$$, so $$g(f(h(x))))=0$$. From here one can conclude $$\psi \in \mathrm{Ker}(g_*)$$.

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

• Note that you also have to show that $f_*$ is injective. Dec 11, 2014 at 14:01
• 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. Dec 11, 2014 at 19:24
• @jflipp Why we needs to show that?
– user1167379
Oct 17, 2023 at 19:33

We can argue as follows. Pick an $$h \in \ker g_* \subseteq \mathrm{Hom}_R(M, M_2)$$. Then we have $$g \circ h = 0.$$ This means $$\newcommand{\im}{\operatorname{im}} \im h \subseteq \ker g = \im f$$. Since $$f$$ is injective, we have the $$R$$-homomorphism $$(f|_{\im f})^{-1} \colon \im f \rightarrow M_1$$. So we can define $$j := (f|_{\im f})^{-1} \circ h \colon M \rightarrow M_1$$. 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.