If $f \colon M \to N$ is surjective, $K \subseteq \ker f \subseteq M$, and $M/K \cong N$, does that imply $K = \ker f$?
Solution 1:
No, this does not hold. Let $R = \mathbb{Z}$.
Consider $M = \bigoplus\limits_{i = 0}^\infty \mathbb{Z}$, generated by $\{z_i\}_{i = 0}^\infty$.
Consider $N = M$.
Consider the map $f : M \to N$ defined by $f(z_0) = 0$ and $f(z_{i + 1}) = z_i$.
Then $f$ is surjective, and its kernel is the copy of $\mathbb{Z}$ generated by $z_0$.
However, let $K = 0$. Then $M / K \cong M \cong N$.
Note that there's nothing special about $\mathbb{Z}$ here - we could have substituted any nonzero ring and gotten this result.