Thread: Proving Functions are Onto

Let f; g : R -> R be two functions. Prove that if both f and g are surjective then the
composition f o g is surjective. Show by means of an example that the converse statement is
not true.

it would really help me if you can post an explantion on how you solved it from the begining to the end.

If f is surjective function, it means 2 things:

1. It has a value for all input points (Because f is a function).
2. It has a input point for any output point (Because f is surjective).

Same things if g is surjective also.

Then, if you take a point p in R, then you can always have a point (at least one) p_1, such that g(p_1) = p (Because g has something for any point p in the co-domain). If we consider the point p_1, there is at least one point p_2 in R such that, f(p_2) = p_1. So, combining these two,

For any point

gof(p_2) = g(f(p_2)) = g(p_1) = p

So, for any point p, there exists (at least) one point p_2, such that, gof(p_2) = p. proves that gof is surjective.

For the other part, it is better to give an example. Consider.

f(x) = ln(abs(x)).
g(x) = e^x.

g is clearly not surjective, because it maps only to positive values in the range. But, fog is surjective. Because,

fog(x) = f(g(x)) = f(e^x) = ln(e^x) = x, which is same as identity, that means, it is surjective.

Salahuddin
Maths online