Let $A$ and $B$ be (associative or Lie) algebras over a field $F$, and let $I \subseteq A$ be an ideal. Suppose we have $A/I \cong B$. Do we then have $A \cong B \oplus I$?
The answer may be different depending on whether the Algebra is associative or is a Lie Algebra (thought I doubt it).
I should point out that this is not homework.
No, a counterexample: take in $F[x]$ the ideal $F[x]x^2$ generated by $x^2$. Then in $F[x]/F[x]x^2$ there are divisors of zero (e.g., $\bar{x}^2=0$ where $\bar{x}$ is the image of $x$), whereas in $F[x]$ they are not.