My question is about exact sequence and tensor product of modules. Consider the following exact sequence of $R$-module for a commutative ring $R$ as in here: \begin{equation} 0 \to N_1 \to N_2 \to N_3 \to 0 \ \ (*)\end{equation} Taking tensor product with $R$-module $M$, we get the exact sequence \begin{equation} M \otimes N_1 \to M \otimes N_2 \to M \otimes N_3 \to 0 \ \ (**)\end{equation} But if $M$ is a flat $R$-module, we get the exact sequence \begin{equation} 0 \to M \otimes N_1 \to M \otimes N_2 \to M \otimes N_3 \to 0 \ \ (***)\end{equation} Why is so ? What is the difference between the exact sequences $(**)$ and $(***)$ ?
What is special about the map $0 \to M \otimes N_1$ ?