[proofplan] We use bijectivity to treat each element of $N$ as having a unique preimage in $M$. To prove that the inverse is $R$-linear, we verify separately that it preserves addition and scalar multiplication. In each case, we apply $f$ to the proposed equality in $M$ and use injectivity, equivalently uniqueness of preimages, to identify the two sides. [/proofplan] [step:Define the inverse map and record uniqueness of preimages] Since $f:M\to N$ is bijective, its inverse function is well-defined. Define \begin{align*} g:N\to M \end{align*} by $g(y)=f^{-1}(y)$ for each $y\in N$. Thus, for every $y\in N$, the element $g(y)\in M$ is the unique element of $M$ satisfying \begin{align*} f(g(y))=y. \end{align*} It is enough to prove that $g$ preserves addition and scalar multiplication. [/step] [step:Use additivity of $f$ to prove additivity of $g$] Let $y_1,y_2\in N$. Define $x_1:=g(y_1)$ and $x_2:=g(y_2)$, so $x_1,x_2\in M$ and \begin{align*} f(x_1)=y_1 \end{align*} and \begin{align*} f(x_2)=y_2. \end{align*} Since $f$ is an $R$-[module homomorphism](/page/Module%20Homomorphism), it is additive. Therefore \begin{align*} f(x_1+x_2)=f(x_1)+f(x_2)=y_1+y_2. \end{align*} By the uniqueness of the preimage of $y_1+y_2$ under the bijection $f$, we obtain \begin{align*} g(y_1+y_2)=x_1+x_2. \end{align*} Substituting $x_1=g(y_1)$ and $x_2=g(y_2)$ gives \begin{align*} g(y_1+y_2)=g(y_1)+g(y_2). \end{align*} [guided] We want to prove that the inverse map respects addition in $N$. Take arbitrary elements $y_1,y_2\in N$. Because $g:N\to M$ is the inverse of $f$, define \begin{align*} x_1:=g(y_1) \end{align*} and \begin{align*} x_2:=g(y_2). \end{align*} Then $x_1,x_2\in M$, and the defining property of the inverse gives \begin{align*} f(x_1)=y_1 \end{align*} and \begin{align*} f(x_2)=y_2. \end{align*} The candidate value for $g(y_1+y_2)$ is $x_1+x_2$, because applying the additive map $f$ to this sum gives \begin{align*} f(x_1+x_2)=f(x_1)+f(x_2)=y_1+y_2. \end{align*} This shows that $x_1+x_2$ is a preimage of $y_1+y_2$ under $f$. Since $f$ is bijective, every element of $N$ has exactly one preimage in $M$. Hence the preimage of $y_1+y_2$ is precisely $x_1+x_2$, so \begin{align*} g(y_1+y_2)=x_1+x_2. \end{align*} Replacing $x_1$ and $x_2$ by their definitions gives \begin{align*} g(y_1+y_2)=g(y_1)+g(y_2). \end{align*} Thus $g$ preserves addition. [/guided] [/step] [step:Use scalar compatibility of $f$ to prove scalar compatibility of $g$] Let $r\in R$ and $y\in N$. Define $x:=g(y)$, so $x\in M$ and \begin{align*} f(x)=y. \end{align*} Since $f$ is an $R$-module homomorphism between left $R$-modules, it satisfies scalar compatibility: \begin{align*} f(rx)=r f(x)=r y. \end{align*} Thus $rx\in M$ is a preimage of $ry\in N$ under $f$. By uniqueness of preimages under the bijection $f$, \begin{align*} g(ry)=rx. \end{align*} Substituting $x=g(y)$ gives \begin{align*} g(ry)=r g(y). \end{align*} [/step] [step:Conclude that the inverse is an $R$-module homomorphism] We have proved that for all $y_1,y_2\in N$, \begin{align*} g(y_1+y_2)=g(y_1)+g(y_2), \end{align*} and for all $r\in R$ and $y\in N$, \begin{align*} g(ry)=r g(y). \end{align*} Therefore $g:N\to M$ is an $R$-module homomorphism. Since $g=f^{-1}$ by definition, the inverse function $f^{-1}:N\to M$ is an $R$-module homomorphism. [/step]