Find $f(0)$

$\begin{aligned} f(x+f(y)-1) & = y + f(x) \quad \quad \color{#3D99F6}{\text{Let } y = x} \\ f(x+f(x)-1) & = x + f(x) \quad \quad \color{#3D99F6}{\text{Let } x + f(x) = z} \\ f(z-1) & = z \quad \quad \quad \quad \quad \ \color{#3D99F6}{\text{Let } z = 1} \\ f(0) & = \boxed{1} \end{aligned}$

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Previous solution

Since $y$ in the RHS of $f(x+f(y)-1) = y + f(x)$ has a degree of 1, for the LHS = RHS for all $x$ and $y$ , the degree of $f$ must be 1. So, we can assume $f(x) = ax + b$ . Then we have:

$\begin{aligned} f(x+f(y)-1) & = y + f(x) \\ f(x + ay + b -1) & = y + ax + b \\ ax + a^2y + ab - a + b & = ax + y + b \\ a^2y + ab - a & = y \end{aligned}$

Equating the coefficients of $y$ and the constant on both sides:

$\implies \begin{cases} a^2 = 1 & \implies a = \pm 1 \\ ab - a = 0 & \implies b = 1 \end{cases}$

$\begin{aligned} \implies f(x) & = \pm x + 1 \\ f(0) & = \boxed{1} \end{aligned}$

$f(x+f(y)-1)=y+f(x)\cdots(1)$
Put $x=1$ , $f(f(y))=y+f(1)\cdots(2)$
Apply $f$ on both sides of $(1)$ and simplify with $(2)$ ,
$f(f(x+f(y)-1))=f(y+f(x))$
$x+f(y)-1+f(1)=f(y+f(x))$
Put $y=0$ , $x+f(0)-1+f(1)=f(f(x))\cdots(3)$
Compare $(2)$ with $(3)$ , $f(0)=1$

Put $y =0$ and $x= 1$

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

$\Rightarrow f({\color{#20A900}{f(0)}}) = f({\color{#20A900}{1}})$

Hence, $f(0) = \boxed{1}$

Taking , $x=1$ and $y=0$ ,

we get , $f(1+f(0)-1)=0+f(1)$ ,

$(f(f(0))) =f(1) \implies f(0) = 1$