Follow on from previous question: Functional Equation - a little tricky

Find all functions $f:\mathbb{R} \rightarrow \mathbb{R}$ such that $f[f(x)^2+f(y)]=xf(x)+y$ 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.

• Why does $f(f(y))=y$ show injectivity? sorry for my naivety. And also why does $f[f(0)^2+f(y)]=y$ show surjectivity? I apologize for my ignorance. – Dominic Stone Apr 12 '13 at 6:58
• @DominicStone go with defition $f(x)=f(y) \Rightarrow f(f(x))=f(f(y))\Rightarrow x=y$. For the surjective part since $f[f(f(0)^2+f(y)]=y$ and $y$ can take any value in $\mathrm{R}$ and by the previous equation $y$ is also in the image of $f$ since it is the image of $f[f(0)^2+f(y)]$ we conclude – clark Apr 12 '13 at 7:17
• Thanks you SO MUCH. I now completely understand. I was being VERY blind. Thanks again! – Dominic Stone Apr 12 '13 at 7:23

Injectivity and surjectivity follow from relatively abstract properties of the form of the equation.

Injectivity follows because for any $x$, the left side (LHS) is a function of $f(y)$ while the right side (RHS) is an injective function of $y$.

Surjectivity because the RHS is a surjective function of $y$, for any $x$, while the LHS is $f(...)$.

Hence the arguments written down in comments under the answer that led to this question:

(to prove surjectivity) adjust $y$ so as to hit any desired value.

The proof of injectivity ... is to compare the functional equation written using $(x,y)$ and $(x,z)$ as the variables. If $f(y)=f(z)$ for fixed $x$ then the left sides of the equations are equal, which compels $y=z$ by looking at the right hand sides.