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

Question

$\newcommand{\Hom}{\mathrm{Hom}}$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?

Answer 1

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 $\varphi'$.