Find all functions $f:\mathbb{R} \to \mathbb{R}$ such that $f\big(f(x)^2+f(y)\big)=xf(x)+y$ for all real numbers $x$ and $y$.
The answer to this has already been posted, but it doesn't explain why this function is injective and surjective. I would really appreciate it if someone did.
Link to that question: Functions satisfying $f\left( f(x)^2+f(y) \right)=xf(x)+y$
Surjective because $f[f(0)^2+f(y)] = y$ for any $y$. In particular, there is an $x_0$ such that $f(x_0)=0$, and using $x=x_0$ in the identity gives $f(f(y))=y$, which shows injectivity.