Determine the functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that:

$$
f(f(x+1)+y-1)=f(x)+y
$$
Let's redo as at the beginning of the course:

1. We try $f(x)=x$ and this function is indeed a solution:

$$
f(f(x+1)+y-1)=f(x+1+y-1)=x+y \quad f(x)+y=x+y
$$

2. We don't know how to get $f(0)$ or another value. Let's skip this for now.
3. Similarly...
4. Is the function surjective? Let $t \in \mathbb{R}$, we choose $x=0$ and $y=f(0)-t$. We then have

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

with $u=f(1)+y-1$ we have $f(u)=t$ so the function is surjective. Since $f$ is surjective, we can choose $x$ to simplify the term $f(x+1)+y-1$. There exists $u$ such that $f(u)=1$ and we set $x_{*}=u-1$ and thus $f\left(x_{*}+1\right)=f(u)=1$. We then have

$$
f(y)=f(1+y-1)=f\left(x_{*}\right)+y
$$

Conclusion: Noting $c=f\left(x_{*}\right)$, we have $f(y)=y+c$. We are almost done; we observe that for $x=0$ and $y=0$, we have $f(1+c+0-1)=c+c=2c$ and $f(0)+0=c$. Therefore, the possible value for $c$ is $c=0$, so $f(y)=y$ for all $y$.