Given that f and g are onto functions, we need to show that f â—¦ g is also onto.
Let y be an arbitrary element in the codomain of f â—¦ g. Since f is onto, there exists an element x in the domain of f such that f(x) = y. Similarly, since g is onto, there exists an element z in the domain of g such that g(z) = x.
Now, consider (f â—¦ g)(z) = f(g(z)) = f(x) = y. This shows that for any y in the codomain of f â—¦ g, there exists a z in the domain of g such that (f â—¦ g)(z) = y. Therefore, f â—¦ g is onto.