[proofplan] We prove both implications directly from the definitions. If $f$ is injective, then any element whose image is $0_N$ has the same image as $0_M$, so it must equal $0_M$. Conversely, if the only element mapped to $0_N$ is $0_M$, then equality of two images forces the difference of the two inputs to lie in the kernel, hence to vanish. [/proofplan] [step:Use injectivity to force every kernel element to be zero] Assume that $f$ is injective. Let $m \in \ker f$. By definition of the kernel, \begin{align*} f(m) = 0_N. \end{align*} Since $f$ is an $R$-module homomorphism, it preserves zero, so \begin{align*} f(0_M) = 0_N. \end{align*} Therefore $f(m)=f(0_M)$. Injectivity of $f$ gives $m=0_M$, so $\ker f \subset \{0_M\}$. Conversely, $0_M \in \ker f$ because $f(0_M)=0_N$, and hence $\{0_M\} \subset \ker f$. Thus \begin{align*} \ker f = \{0_M\}. \end{align*} [/step] [step:Use the zero kernel to force equal preimages to coincide] Assume that \begin{align*} \ker f = \{0_M\}. \end{align*} Let $x,y \in M$ satisfy $f(x)=f(y)$. Since $f$ is an $R$-module homomorphism, it is additive, and therefore \begin{align*} f(x-y)=f(x)-f(y)=0_N. \end{align*} Hence $x-y \in \ker f$. By the assumption on the kernel, $x-y=0_M$. Adding $y$ to both sides in the underlying abelian group of $M$ gives $x=y$. Therefore $f$ is injective. [guided] Assume that the kernel is exactly the zero submodule: \begin{align*} \ker f = \{0_M\}. \end{align*} To prove that $f$ is injective, we must show that any two elements of $M$ with the same image under $f$ are equal. Let $x,y \in M$ and assume \begin{align*} f(x)=f(y). \end{align*} The natural object to examine is the difference $x-y$, because additivity of the module homomorphism converts equality of images into membership in the kernel. Since $f$ is additive, \begin{align*} f(x-y)=f(x)+f(-y). \end{align*} An additive homomorphism satisfies $f(-y)=-f(y)$, because $0_N=f(0_M)=f(y+(-y))=f(y)+f(-y)$. Hence \begin{align*} f(x-y)=f(x)-f(y)=0_N. \end{align*} By definition of the kernel, this means $x-y \in \ker f$. The hypothesis $\ker f=\{0_M\}$ now forces \begin{align*} x-y=0_M. \end{align*} Adding $y$ to both sides in the additive group of the module $M$ yields $x=y$. Thus whenever $f(x)=f(y)$, we have $x=y$, which is precisely injectivity of $f$. [/guided] [/step] [step:Combine the two implications] The first step proves that injectivity of $f$ implies $\ker f=\{0_M\}$. The second step proves that $\ker f=\{0_M\}$ implies injectivity of $f$. Therefore $f$ is injective if and only if \begin{align*} \ker f = \{0_M\}. \end{align*} [/step]