$f(f(y)+1)=y+f(1)$ is bijective.

Question

Find all functions $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $$f(xf(y)+x)=xy+f(x), \; \forall x,y \in \mathbb{R}.$$

I read a solution in finding this function. It states that setting $x=1$ gives $$f\big(f(y)+1\big)=y+f(1)\,,$$ which means $f$ is bijective.

Why does $f\big(f(y)+1\big)=y+f(1)$ imply that $f$ is bijective?

Answer 1

Assume $f(y_1)=f(y_2)$. Then $y_1+f(1)=f(f(y_1)+1)=f(f(y_2)+1)=y_2+f(1)$ and hence $y_1=y_2$. Hence $f$ is injective.

Given $a\in \mathbb R$, let $y=a-f(1)$. Then $f(f(y)+1)=a$. Hence $f$ is surjective.

Answer 2

Because $f(1)$ is a constant, and all functions in the form $y=x+c$ with $c$ a constant are bijective.

Therefore $y+f(1)$ is bijective, thus $f(f(y)+1)$ is bijective.

Form this, we may conclude that $f(x)$ is surjective and $f(x)+1$ is injective, so $f(x)$ is surjective and injective, so $f(x)$ is bijective.

Answer 3

Below is a solution to this functional equation.

As stated before, $f$ is bijective. Hence, there exist $a,b\in\mathbb{R}$ such that $f(a)=0$ and $f(b)=-1$. Replace $x,y$ by $a$ to get $$0=f(a)=f\big(a\cdot f(a)+a\big)=a\cdot a+f(a)=a^2\,.$$ Hence, $a=0$, or $f(0)=0$. Replace $y$ by $b$ to get $$0=f(0)=f\big(x\cdot f(b)+x\big)=x\cdot b+f(x)\,.$$ Ergo, $f(x)=-bx$ for all $x\in\mathbb{R}$. Plugging this into the original functional equation, we obtain $$b^2xy-bx=-b\big(x(-by)+x\big)=f\big(x\cdot f(y)+x\big)=xy+f(x)=xy-bx\,.$$ That is, $b^2=1$, or $b=-1$ or $b=+1$. Ergo, either $$f(x)=-x\text{ for all }x\in\mathbb{R}$$ or $$f(x)=+x\text{ for all }x\in\mathbb{R}\,.$$ Both functions are indeed solutions, so we are done.