# Proof that $0 \to \mathrm{Hom}(L,D) \overset{\varphi'}{\to} \mathrm{Hom}(N,D) \overset{\psi '}{\to} \mathrm{Hom}(M,D)$ is an exact sequence

$$\newcommand{\Hom}{\mathrm{Hom}}$$Let $$M,N,L$$ and $$D$$ modules over a ring $$R$$. Let $$Hom(M,D)$$ denote the $$R-$$module of the homomorphisms $$M \to D$$. Given an homomorphism $$\varphi: M \to N$$ we define $$\varphi ': \Hom(N,D) \to \Hom(M,D)$$ by $$\varphi'(f) = f \circ \varphi.$$

Now, suppose that $$0 \to M\overset{\psi}{\rightarrow} N \overset{\varphi}{\rightarrow} L \to 0$$ an exact short sequence, i.e $$\psi$$ is injective, $$\varphi$$ is surjective and $$\ker \varphi = \mathrm{im}\, \psi$$ I'm trying to proof that $$0 \to \Hom(L,D) \overset{\varphi '}{\rightarrow} \Hom(N,D) \overset{\psi '}{\rightarrow} \Hom(M,D)$$ is an exact sequence, i.e $$\varphi '$$ is injective and $$\ker \psi ' = \mathrm{im}\, \varphi '$$.

My attempt: Since $$\varphi$$ is surjective, $$f_1 \circ \varphi = f_2 \circ \varphi \iff f_1 = f_2$$, hence $$\varphi '$$ is injective. If $$f \in im \, \varphi '$$, exists $$g \in \Hom(L,D), \, g \circ \varphi = f \Rightarrow f \circ \psi = g \circ (\varphi \circ \psi) = 0 \Rightarrow f \in \ker \psi '.$$

However, I couldn't proof the inclusion $$\ker \psi ' \subset \mathrm{im} \, \varphi '.$$

Help?

Think of $M$ as a submodule of $N$. Then $\ker\psi'$ consists of the homomorphisms $f:N\to D$ which when restricted to $M$ are zero, that is we require $f(m)=0$ for all $m\in M$. Such a homomorphism induces a homomorphism from $N/M$ to $D$, that is it is effectively in $\textrm{Hom}(N/M,D)\cong \textrm{Hom}(L,D)$, that is in the image of $\varphi'$.
• Why can we think of $M$ as a submodule of $N$? Commented Jan 1, 2018 at 23:19