Prove Set of all onto mappings from A->A is closed

[knowLittle]

Homework Statement

Prove that set of all onto mappings of A->A is closed under composition of mappings:

Homework Equations

Definition of onto and closure on sets.

The Attempt at a Solution

Say, $f$ and $g$ are onto mappings from A to A.
Now, say I have a set S(A) = {all onto mappings of A to A }, so $f$ and $g \in S$
Then,

$f o g(a_0) = f(g(a_0)) = f(a_i) , a_i, a_0 \in A; a_i$ represents all elements in A that are being hitted.
Now, the domain of $f$ is set A, since $g$ is onto. And now all elements in the domain of $f$ will hit all elements in the codomain of f.
Therefore, $fog \in S(A)$

Is this correct? Or is it weak, illogical, flawed ... ?

 

[PeroK]

Homework Statement

Prove that set of all onto mappings of A->A is closed under composition of mappings:

Homework Equations

Definition of onto and closure on sets.

The Attempt at a Solution

Say, $f$ and $g$ are onto mappings from A to A.
Now, say I have a set S(A) = {all onto mappings of A to A }, so $f$ and $g \in S$
Then,

$f o g(a_0) = f(g(a_0)) = f(a_i) , a_i, a_0 \in A; a_i$ represents all elements in A that are being hitted.
Now, the domain of $f$ is set A, since $g$ is onto. And now all elements in the domain of $f$ will hit all elements in the codomain of f.
Therefore, $fog \in S(A)$

Is this correct? Or is it weak, illogical, flawed ... ?

It's not correct. Both the general approach and the details are not right.

First, there's no need to introduce S(A) when you did. Stick with f and g being onto. You need to show that f o g is onto. Now, can you write down the definition of onto for f o g? Try writing it as:

"We need to show that ..."

 

[knowLittle]

Thanks.
We need to show that $fog$ is onto or in other words
fog = z = f(g(x)) =A?

f: A-> A : f(a) = A
g: A ->A: g(a) = A

Is this correct?

 

[SammyS]

Thanks.
We need to show that $fog$ is onto or in other words
fog = z = f(g(x)) =A?

f: A-> A : f(a) = A
g: A ->A: g(a) = A

Is this correct?

You need to show that for every z in A, there is an x in A such that $\ (f \circ g) \, (x)=z \$ .

 

[knowLittle]

Still not very sure on how to start.
BTW, is there anything rescuable from my first approach?

 

[SammyS]

Still not very sure on how to start.
BTW, is there anything rescuable from my first approach?

Sure.

Having f & g be chosen from set S is fine.

Then show that $\ f\circ g\$ is onto.

 

[PeroK]

Thanks.
We need to show that $fog$ is onto or in other words
fog = z = f(g(x)) =A?

f: A-> A : f(a) = A
g: A ->A: g(a) = A

Is this correct?

If you don't know the definition of onto, how can you prove a function is onto? Formal definitions are essential for rigorous mathematics.

Does the formal definition that Sammy gave you make sense?

I'd do two things to start. First, I'd write down what you know about f and g being onto.

Then, I'd find an example. Perhaps f, g: [0, 1] -> [0, 1], where f(x) = x^2 and g(x) = 1 - x. Check that these are onto, then show that f o g is onto. Why is f o g onto? Then, try to generalise this for any functions.