# If g(f(x)) is one-to-one (injective) show f(x) is also one-to-one (given that…)

Possible Duplicate:
Injective and Surjective Functions

If $g(f(x))$ is one-to-one (injective) show $f(x)$ is also one-to-one given that $f$ is a function from $A$ to $B$ and $g$ a function from $B$ to $C$.

I’ve just started my Discrete math course and I’d like some help on this. I’m pretty sure we’re supposed to use set theory laws to prove this.

So far I know the three conditions that satisfy an injective function (sorry, having difficulties typing all this TeX markup so I’ll skip that).

Any help?

#### Solutions Collecting From Web of "If g(f(x)) is one-to-one (injective) show f(x) is also one-to-one (given that…)"

Hint: suppose that $f$ is not one to one. Then there are $x \ne y$ such that $f(x)=f(y)$. Can you conculude that $g\circ f$ is not one to one?

Or the other way around:
Take $x \ne y$. Since $g \circ f$ is injective we have $g(f(x))\ne g(f(y))$, so we must have $f(x)\ne f(y)$.

Hence $g \circ f$ is injective then by definition it means that for all $x, y$ in the domain of $f$ (being the domain of $g \circ f$) we get
$$[(g \circ f) (x) = (g \circ f)(y)] \Rightarrow x = y.$$
Assume now that $f(x) = f(y)$. Then of course $(g \circ f)(x) = g(f(x)) = g(f(y)) = (g \circ f)(y)$. By previous implication we obtain $x = y$ and hence $f$ is injective.