Find the functions $f$ : $\mathbb{R} \rightarrow \mathbb{R}$ such that for all real numbers $x, y$, we have

$$
f(x-f(y))=1-x-y
$$
By setting $x=f(y)$, we find $f(y)=1-f(0)-y$. Let $c=1-f(0)$, in the initial equation, we have $1-x-y=f(x+y-c)=-(x+y-c)+c=2 c-x-y$ thus $c=\frac{1}{2}$. The only possible solution is therefore $x \mapsto \frac{1}{2}-x$, which indeed satisfies the initial equation.