[proofplan] Define the composite map $h=g\circ f$. Since $f$ maps $L$ into $M$ and $g$ maps $M$ into $N$, this composite is a well-defined function from $L$ to $N$. We then verify the two defining properties of an $R$-[module homomorphism](/page/Module%20Homomorphism): preservation of addition and compatibility with scalar multiplication. [/proofplan] [step:Define the composite as a map from $L$ to $N$] Define \begin{align*} h:L &\to N \end{align*} by \begin{align*} h(x)=g(f(x)) \end{align*} for each $x\in L$. This is well-defined because $f(x)\in M$ for every $x\in L$, and $g$ has domain $M$. [/step] [step:Verify that the composite preserves addition] Let $x,y\in L$. Since $f$ is an $R$-module homomorphism, it is additive, so \begin{align*} f(x+y)=f(x)+f(y). \end{align*} Applying the additivity of the $R$-module homomorphism $g:M\to N$, we obtain \begin{align*} h(x+y)=g(f(x+y)). \end{align*} Substituting the additivity identity for $f$ gives \begin{align*} h(x+y)=g(f(x)+f(y)). \end{align*} Using additivity of $g$ gives \begin{align*} h(x+y)=g(f(x))+g(f(y)). \end{align*} By the definition of $h$, this is \begin{align*} h(x+y)=h(x)+h(y). \end{align*} [guided] We must show that the composite map preserves the addition operation on the source module $L$. Choose arbitrary elements $x,y\in L$. Because $f:L\to M$ is an $R$-module homomorphism, it preserves addition in the sense that \begin{align*} f(x+y)=f(x)+f(y), \end{align*} where the addition on the left is the addition in $L$, and the addition on the right is the addition in $M$. Now apply the definition of the composite map $h=g\circ f$. We have \begin{align*} h(x+y)=g(f(x+y)). \end{align*} Using the additivity of $f$, this becomes \begin{align*} h(x+y)=g(f(x)+f(y)). \end{align*} The elements $f(x)$ and $f(y)$ both lie in $M$, so the additivity of the homomorphism $g:M\to N$ applies to their sum. Hence \begin{align*} g(f(x)+f(y))=g(f(x))+g(f(y)). \end{align*} Combining these equalities and using $h(x)=g(f(x))$ and $h(y)=g(f(y))$, we get \begin{align*} h(x+y)=h(x)+h(y). \end{align*} Thus $h$ preserves addition. [/guided] [/step] [step:Verify that the composite is compatible with scalar multiplication] Let $r\in R$ and $x\in L$. Since $f$ is an $R$-module homomorphism, it is compatible with the left $R$-actions, so \begin{align*} f(rx)=r f(x). \end{align*} Using the definition of $h$, we have \begin{align*} h(rx)=g(f(rx)). \end{align*} Substituting the scalar-compatibility identity for $f$ gives \begin{align*} h(rx)=g(r f(x)). \end{align*} Since $g$ is an $R$-module homomorphism, it is compatible with scalar multiplication, so \begin{align*} g(r f(x))=r g(f(x)). \end{align*} Therefore \begin{align*} h(rx)=r h(x). \end{align*} [/step] [step:Conclude that the composite is an $R$-module homomorphism] We have shown that $h:L\to N$ preserves addition and is compatible with scalar multiplication by every $r\in R$. By the definition of an $R$-module homomorphism, $h$ is an $R$-module homomorphism. Since $h=g\circ f$, the composite $g\circ f:L\to N$ is an $R$-module homomorphism. [/step]