[proofplan]
We use pointwise operations. First, pointwise addition gives $\operatorname{Hom}_R(M,N)$ an abelian group structure. Then we check that the scalar multiple $rf$ of an $R$-[linear map](/page/Linear%20Map) is again $R$-linear; the only non-formal point is where commutativity of $R$ is needed to move the scalar $r$ past another scalar. Finally, all scalar multiplication axioms are verified by evaluating at an arbitrary element of $M$ and using the corresponding module axiom in $N$.
[/proofplan]
[step:Use pointwise addition to obtain the underlying abelian group]
By [citetheorem:8335], applied to the ring $R$ and the left $R$-modules $M$ and $N$, the set $\operatorname{Hom}_R(M,N)$ is an abelian group under pointwise addition. Explicitly, the addition is the operation
\begin{align*}
(f+g)(m)=f(m)+g(m)
\end{align*}
for $f,g\in \operatorname{Hom}_R(M,N)$ and $m\in M$, the zero element is the zero homomorphism $0_{\operatorname{Hom}}:M\to N$ given by $0_{\operatorname{Hom}}(m)=0_N$, and the additive inverse of $f$ is the map $-f:M\to N$ given by $(-f)(m)=-f(m)$.
[/step]
[step:Show that scalar multiplication is closed on $\operatorname{Hom}_R(M,N)$]
Fix $r\in R$ and $f\in \operatorname{Hom}_R(M,N)$. Define the function
\begin{align*}
rf:M&\to N
\end{align*}
by
\begin{align*}
(rf)(m)=r f(m)
\end{align*}
for every $m\in M$. We prove that $rf$ is an $R$-[module homomorphism](/page/Module%20Homomorphism).
Let $m_1,m_2\in M$. Since $f$ is additive and the scalar action of $R$ on $N$ distributes over addition in $N$,
\begin{align*}
(rf)(m_1+m_2)=r f(m_1+m_2)=r(f(m_1)+f(m_2))=r f(m_1)+r f(m_2)=(rf)(m_1)+(rf)(m_2).
\end{align*}
Let $s\in R$ and $m\in M$. Since $f$ is $R$-linear, the scalar action on $N$ is associative, and $R$ is commutative,
\begin{align*}
(rf)(sm)=r f(sm)=r(s f(m))=(rs)f(m)=(sr)f(m)=s(r f(m))=s(rf)(m).
\end{align*}
Thus $rf$ is additive and respects scalar multiplication, so $rf\in \operatorname{Hom}_R(M,N)$.
[guided]
Fix $r\in R$ and $f\in \operatorname{Hom}_R(M,N)$. The scalar action in the statement defines a function
\begin{align*}
rf:M&\to N
\end{align*}
by
\begin{align*}
(rf)(m)=r f(m)
\end{align*}
for every $m\in M$. To use this as scalar multiplication on $\operatorname{Hom}_R(M,N)$, we must first prove that this function is still an element of $\operatorname{Hom}_R(M,N)$; that is, it must be an $R$-module homomorphism.
First check additivity. Let $m_1,m_2\in M$. Because $f$ is an $R$-module homomorphism, it is additive, so $f(m_1+m_2)=f(m_1)+f(m_2)$. Then the distributive law for the scalar action on the module $N$ gives
\begin{align*}
(rf)(m_1+m_2)=r f(m_1+m_2)=r(f(m_1)+f(m_2))=r f(m_1)+r f(m_2)=(rf)(m_1)+(rf)(m_2).
\end{align*}
Now check compatibility with scalar multiplication. Let $s\in R$ and $m\in M$. Since $f$ is $R$-linear, $f(sm)=s f(m)$. Therefore
\begin{align*}
(rf)(sm)=r f(sm)=r(s f(m)).
\end{align*}
The module axiom in $N$ allows us to combine successive scalar actions:
\begin{align*}
r(s f(m))=(rs)f(m).
\end{align*}
Here is the one place where commutativity of $R$ is essential. Since $R$ is commutative, $rs=sr$, and hence
\begin{align*}
(rs)f(m)=(sr)f(m)=s(r f(m))=s(rf)(m).
\end{align*}
Thus
\begin{align*}
(rf)(sm)=s(rf)(m).
\end{align*}
We have proved that $rf$ is additive and $R$-linear, so $rf\in \operatorname{Hom}_R(M,N)$.
[/guided]
[/step]
[step:Verify the scalar multiplication axioms pointwise]
It remains to verify the module axioms involving scalar multiplication. Let $r,s\in R$ and $f,g\in \operatorname{Hom}_R(M,N)$. Since equality of functions $M\to N$ is pointwise equality, it suffices to evaluate each identity at an arbitrary $m\in M$.
For compatibility of scalar multiplication with multiplication in $R$, the associativity axiom for the $R$-module $N$ gives
\begin{align*}
((rs)f)(m)=(rs)f(m)=r(s f(m))=(r(sf))(m).
\end{align*}
Thus $(rs)f=r(sf)$.
For the identity scalar, the identity axiom for the $R$-module $N$ gives
\begin{align*}
(1_R f)(m)=1_R f(m)=f(m).
\end{align*}
Thus $1_R f=f$.
For distributivity over addition in the ring variable, the distributive axiom for the scalar action on $N$ gives
\begin{align*}
((r+s)f)(m)=(r+s)f(m)=r f(m)+s f(m)=(rf)(m)+(sf)(m)=((rf)+(sf))(m).
\end{align*}
Thus $(r+s)f=rf+sf$.
For distributivity over addition in the module variable, the distributive axiom for the scalar action on $N$ gives
\begin{align*}
(r(f+g))(m)=r((f+g)(m))=r(f(m)+g(m))=r f(m)+r g(m)=(rf)(m)+(rg)(m)=((rf)+(rg))(m).
\end{align*}
Thus $r(f+g)=rf+rg$.
[/step]
[step:Conclude that $\operatorname{Hom}_R(M,N)$ is an $R$-module]
The first step gives an abelian group structure on $\operatorname{Hom}_R(M,N)$ under pointwise addition. The second step shows that the prescribed scalar multiplication is a well-defined map
\begin{align*}
R\times \operatorname{Hom}_R(M,N)&\to \operatorname{Hom}_R(M,N).
\end{align*}
The third step verifies the remaining scalar multiplication axioms. Therefore $\operatorname{Hom}_R(M,N)$ is a left $R$-module under pointwise addition and the scalar multiplication $(rf)(m)=r f(m)$.
[/step]
