$f(yf(x) + y) = xy + f(y)$ for all $x, y.$ Prove $f$ is surjective

Question

I'm stucking with this problem for days. f is a function from $\mathbb{R}$ to $\mathbb{R}$ satisfies:

$$f(yf(x) + y) = xy + f(y) \qquad \forall x, y \in \mathbb{R}$$

Prove $f$ is surjective, and hence find all the functions satisfy this equation.

I think fixing a $x$ and $y$ as a constant and study the function might be a good idea, but so far I've been only playing with $x=1,0$ and $y=1,0,$ and they didn't seem to help in proving surjectivity.

Any help is appreciated. Thanks!

(Proving injectivity is easy)

Answer 1

First, notice that by defining$ g(x,y)=yf(x)+y$ , and using the injectivity you have found, then $f \circ g(0,y) $, you can find out that f(0)=0.
Now , find (1,y) s.t $f \circ g (1,y)$ isnt0 ,say C.
Suppose negatively that there is real number $M_0$ s.t $ M_0 $ isn't in Im(f).
$ f \circ g (1,y) = f(y) +y $
Notice that in this case (by the given equality) , for any $ y \in \mathbb{R} $, $M_0 +y $ isn't in Im(f)- for contradiction, take $y=c-M_0$

Answer 2

Lemma (1): $f(f(x)+1) = x+f(1)$.

Proof of Lemma (1) : let $y = 1$.

Lemma (2): $f$ is injective.

Proof of Lemma (2) by contradiction: if $x \neq y$ and $f(x) = f(y)$, then

$f(f(x)+1) = f(f(y)+1) = x = y$, which contradicts with $x \neq y$.

Lemma (3): $f$ is surjective:

Proof of Lemma (3): By Lemma (1), $f(f(x-f(1))+1) = x - f(1) + f(1) = x$.

Lemma (4): $f(0) = 0$.

Proof of Lemma (4): By Lemma (1):

$f(f(0)+1) = f(1)$.

By the injectivity of $f$, $f(0) + 1 = 1$, then $f(0) = 0$.

I stuck at proving $f(1) = 1$...

Answer 3

Let $f(1)=a$. Then setting $y=1$, we get $$ f\big(f(x) + 1\big) = x + a, \quad \text{for all $x\in\mathbb R$}, $$ which implies that $$ f\big(f(x-a) + 1\big) = x, \quad \text{for all $x\in\mathbb R$}. $$