Check to see if the function $f$ which satisfies the following relation is one-to-one or not: $\forall x,y\in \Bbb R: f(x+f(y))=f(x)+y$

Question

Check to see if the function $f$ which satisfies the following relation is one-to-one or not:
$$\forall x,y\in \Bbb R: f(x+f(y))=f(x)+y$$

Please judge my solution:
I set $y=0$ to get: $f(x+f(0))=f(x)$, now I deduce that this function is one to one if $f(0)=0$, otherwise, for example, $5+f(0)\ne 5$ but $f(x+f(0))=f(x)$. Am I correct??

Answer 1

If $f:\mathbb R\to \mathbb R$ satisfies $f(x+f(y))=f(x)+y$, $\forall x,y\in\mathbb R$, then I'll prove that $f$ is injective.

Let $P(x,y)$ be the statement $f(x+f(y))=f(x)+y$.

Let $f(0)=c$.

$P(-c,0)\implies c=f(-c)$

$P(-c,-c)\implies c=0$

$P(0,y)\implies f(f(y))=y$, $\forall y\in\mathbb R$

If $f(a)=f(b)$, then $f(f(a))=f(f(b))$, so $a=b$. So $f$ is injective.

Answer 2

This is incorrect; rather, the conclusion is the converse of what you said. If $f(0)\neq 0$, then you are correct that $f$ will not be one-to-one. However, if $f(0)=0$, then you cannot conclude that $f$ is one-to-one (all you know is that one possible approach to showing it is not one-to-one fails).

Answer 3

Hint: next assume $f(0)=0$, and see what you get by taking $x=0$.