# Prove that f is injective. (Is my solution correct?)

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
• Doesn't $f$ need to be injective in order to say that $f(f(x1)) = f(f(x2))$ if $f(x1) = f(x2)$? Aug 4, 2015 at 19:47
• "mapping" just means that $f$ is essentially a relation between two sets (domain and range). $f:A\rightarrow B$ means that "$f$ maps $A$ to $B$, i.e. that $f$ takes elements in the set $A$ and "pairs" with elements in the set $B$ but isn't necessarily a function. Aug 4, 2015 at 20:18
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 $x=y$.
• Do you (maybe) mean $f(x) = f(y)$? Aug 4, 2015 at 19:50
• The $y$ you have set is different than mine, isn't it? Cause otherwise you would get $f(x) = f(f(f(x)))$ Aug 4, 2015 at 19:59