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Let $f: R\to R$, and ($f\circ f\circ f)(x) = (f\circ f)(x) + x$, $x$ $\in \mathbb R.$
Prove that $f$ is injective.
My Solution:
Let $x_1, x_2\in \mathbb R.$ and $f(f(x_1)) = f(f(x_2)) = y$
($f\circ f\circ f)(x) = (f\circ f)(x) + x$
$f((f\circ f)(x)) = f(f(x)) + x$
$f(f(f(x))) = f(f(x)) + x$
$f(f(x)) = f(f(f(x))) - x$
Since $f(f(x_1)) = f(f(x_2))$
$f(f(f(x_1))) - x_1 = f(f(f(x_2))) - x_2$
$f(y) - x_1 = f(y) - x_2$
$-x_1 = -x_2$
$x_1 = x_2$ therefore $f$ is injective.
Is this solution acceptable?
You should prove that $f(x_1) = f(x_2)$ implies $x_1 = x_2$. So your solution is not acceptable. You should start with $f(x_1) = f(x_2)$ then say this implies $f(f(x_1)) = f(f(x_2))$ and then complete as you did
It looks on the right path if not a bit long: Suppose that $f(x) = f(y)$. Then $$f(f(x)) = f(f(y))\quad \text{and} \quad f(f(f(x))) = f(f(f(y))).$$ Subtract the second equation from the third to get $x=y$.
