[proofplan] Define the candidate induced map by sending a coset $m+K$ to $f(m)$. The only possible issue is whether this value depends on the chosen representative; the hypothesis $K\subset\ker f$ is exactly what makes $f$ constant on cosets modulo $K$. After well-definedness is established, the [module homomorphism](/page/Module%20Homomorphism) property follows from the quotient operations and the $R$-linearity of $f$. Finally, the factorization is checked on elements of $M$, and uniqueness follows because every element of $M/K$ is the image under $\pi$ of some element of $M$. [/proofplan] [step:Define the induced map on cosets] Define a function $\overline{f}:M/K\to N$ by the rule $\overline{f}(m+K)=f(m)$ for each $m\in M$, provided this rule is well-defined. Here $m+K$ denotes the coset of $m$ in the quotient module $M/K$. [/step] [step:Prove that the value of the induced map does not depend on the representative] Let $m,m'\in M$ satisfy $m+K=m'+K$. By equality of cosets in the quotient module, this means $m-m'\in K$. Since $K\subset\ker f$, we have $m-m'\in\ker f$, and hence \begin{align*} f(m-m')=0_N. \end{align*} Using additivity of the module homomorphism $f$, we obtain \begin{align*} f(m)-f(m')=f(m-m')=0_N. \end{align*} Therefore $f(m)=f(m')$, so the formula $\overline{f}(m+K)=f(m)$ defines a well-defined function $\overline{f}:M/K\to N$. [guided] We must check well-definedness because an element of $M/K$ is a coset, not a chosen representative. Suppose $m,m'\in M$ are two representatives of the same coset, so \begin{align*} m+K=m'+K. \end{align*} By the definition of equality in a quotient module, this is equivalent to \begin{align*} m-m'\in K. \end{align*} The hypothesis $K\subset\ker f$ now gives $m-m'\in\ker f$. By the definition of the kernel of $f$, this means \begin{align*} f(m-m')=0_N. \end{align*} Since $f:M\to N$ is an $R$-module homomorphism, it is a homomorphism of the underlying additive groups, so \begin{align*} f(m-m')=f(m)-f(m'). \end{align*} Combining these two equalities gives \begin{align*} f(m)-f(m')=0_N. \end{align*} Thus $f(m)=f(m')$. Therefore every representative of the same coset gives the same value in $N$, and the rule \begin{align*} \overline{f}(m+K)=f(m) \end{align*} defines a genuine function $\overline{f}:M/K\to N$. [/guided] [/step] [step:Verify that the induced map is an $R$-module homomorphism] We use the [linearity criterion for module homomorphisms](/theorems/8331) [citetheorem:8331]. Let $m_1,m_2\in M$ and let $r_1,r_2\in R$. In the quotient module $M/K$, the module operations are defined by \begin{align*} r_1(m_1+K)+r_2(m_2+K)=(r_1m_1+r_2m_2)+K. \end{align*} Using the definition of $\overline{f}$ and the $R$-linearity of $f$, we compute \begin{align*} \overline{f}(r_1(m_1+K)+r_2(m_2+K))=f(r_1m_1+r_2m_2). \end{align*} Since $f$ is an $R$-module homomorphism, \begin{align*} f(r_1m_1+r_2m_2)=r_1f(m_1)+r_2f(m_2). \end{align*} Using the definition of $\overline{f}$ again, this becomes \begin{align*} \overline{f}(r_1(m_1+K)+r_2(m_2+K))=r_1\overline{f}(m_1+K)+r_2\overline{f}(m_2+K). \end{align*} By the linearity criterion for module homomorphisms [citetheorem:8331], $\overline{f}:M/K\to N$ is an $R$-module homomorphism. [/step] [step:Check that the induced map factors the original homomorphism through the quotient map] For each $m\in M$, the quotient homomorphism satisfies $\pi(m)=m+K$. Therefore \begin{align*} (\overline{f}\circ \pi)(m)=\overline{f}(m+K). \end{align*} By the definition of $\overline{f}$, \begin{align*} \overline{f}(m+K)=f(m). \end{align*} Thus $(\overline{f}\circ \pi)(m)=f(m)$ for every $m\in M$, so \begin{align*} f=\overline{f}\circ \pi. \end{align*} [/step] [step:Use surjectivity of the quotient map to prove uniqueness] Let \begin{align*} g: M/K \to N \end{align*} be an $R$-module homomorphism such that $f=g\circ\pi$. We prove $g=\overline{f}$. Let $x\in M/K$. By the definition of the quotient module, there exists $m\in M$ such that $x=m+K=\pi(m)$. Since $f=g\circ\pi$, we have \begin{align*} g(x)=g(\pi(m))=f(m). \end{align*} By the definition of $\overline{f}$, we also have \begin{align*} \overline{f}(x)=\overline{f}(m+K)=f(m). \end{align*} Therefore $g(x)=\overline{f}(x)$ for every $x\in M/K$, so $g=\overline{f}$. This proves uniqueness and completes the proof. [/step]