Quote Originally Posted by TheMathsDude View Post
Finding it hard to manipulate the surjective definition to fit the previous examples written.

Any clues?
i'll help you out, i leave the actual proof to you, but i'll give you somewhat of an outline.

Definition: A function $\displaystyle f$ is surjective (or onto) if for every $\displaystyle y$ in the codomain of $\displaystyle f$, there is an $\displaystyle x$ in the domain of $\displaystyle f$ such that $\displaystyle f(x) = y$

now what that basically means, is that the range of the function is the same as the codomain of the function.

so we have $\displaystyle f : A \to B$ and $\displaystyle g:B \to C$. so $\displaystyle g \circ f : A \to C$.

so, since $\displaystyle g \circ f$ is surjective, $\displaystyle C$ is the codomain and at the same time, the range of $\displaystyle g \circ f$. now what $\displaystyle g \circ f$ does, is use the function $\displaystyle f$ to take some element $\displaystyle x \in A$ and put it in $\displaystyle B$ and then use the function $\displaystyle g$ to take that element and then move it to $\displaystyle C$. this last part of the function is what makes it surjective, since it is the part that maps $\displaystyle x$ to the range (which is the codomain).