Little confused about the constraint of Injective Functions and Surjective.

Question

From my understanding,

A Function is called to be Injective, if different elements of the first set are mapped to different elements of the second set.

Let set A = {a,b,c} and set B = {1,2,3}

Are the following points correct,

if:

1) f(a) = 1 , f(b) = 2 , f(c) = 3 > The function F: A -> B is injective.

2) f(a) = 1 , f(b) = 1 , f(c) = 3 > The function F: A -> B is not injective.

Did I cover all the possible possibilities?

A Function is called to be Surjective, if all elements of the second set are mapped to some element of the first set.

Let set A = {a,b,c, d, e} and set B = {1,2,3}

Are the following points correct,

if:

1) f(a) = 1 , f(b) = 2 , f(c) = 3 > The function F: A -> B is surjective.

2) f(a) = 1 , f(b) = 1 , f(c) = 3 > The function F: A -> B is not surjective.

Did I cover all the possible possibilities?

Thank you!

Answer 1

Your definitions are wrong.

Your second example says $f(a) = 1$, $f(a) = 2$, $f(c) = 3$. That is not even a function. $f(a)$ cannot be more than one thing.

"Injective" means no two elements of the domain are mapped to the same element of the codomain.

"Surjective" means for every element of the codomain, some element of the domain is mapped to it.

Answer 2

Yes, this is all correct. Of course you can construct more functions $f: A \to B$, that are not injective / not surjective.

Your definition of surjective sounds a little strange, but I think you mean the right thing. A function $f: A \to B$ is called surjective, if for every element $b \in B$ there exists an element $a \in A$, such that $f(a) = b$.