MHB Is Every Statement Correct in Defining an Onto Function?

[lemonthree]

Which of the following statements is equivalent to saying that a function f:A→B is onto? There are 8 options, select all that are correct.

	x	f(A)=B
	x	In the arrow diagram representing f, every point in B has an arrow pointing at it.
	x	∀y∈B ∃x∈A such that f(x)=y
	x	f−1(B)=A
	x	Every element of B is the image of some element in A
		∀x∈A ∃y∈B such that f(x)=y
		Every element of A has a corresponding image in B
	x	Every element of B has at least one preimage.

I have marked with an "x" those that I believe are correct. However, I am not entirely sure.
Let me explain my selection.
1. f(A) = B is true, because B is the whole set B where A is mapped to B, so it is onto? (range = codomain)
2. Valid, because it shows that the range = codomain.
3. Similar to point 2, it shows that range = codomain.
4. In this case, it is similar to 1? It seems to say that the whole set of B can be mapped back to A.
5. Yes, every element of B (codomain) is the image of some x in A.
6. Not true for onto function, because it does not cover all of B, only some in B.
7. Not true for onto function, because it does not talk about covering all options in B.
8. True because every element of B is covered.

Please point out if I have made any mistakes in my explanation, thank you!

 

[HOI]

The definition of "function from A to B", whether "onto" or not, requires that
"∀x∈A ∃y∈B such that f(x)=y".

 

[lemonthree]

Indeed, I agree with you. However, I did not select it because the question asks "Which of the following statements is equivalent to saying that a function f:A→B is onto?". So I'm guessing that a normal function won't make the cut?

 

[castor28]

Regarding point 4, $f^{-1}(B) = \{x \in A \mid f(x)\in B\}$. Since B is the co-domain, this is true for any function $f:A\to B$.

 

[lemonthree]

		f(A)=B
	x	In the arrow diagram representing f, every point in B has an arrow pointing at it.
	x	∀y∈B ∃x∈A such that f(x)=y
		f−1(B)=A
	x	Every element of B is the image of some element in A
		∀x∈A ∃y∈B such that f(x)=y
		Every element of A has a corresponding image in B
	x	Every element of B has at least one preimage.

Thank you very much @castor28 for pointing that out. Following that logic, I should also not select f(A)=B because B is the codomain (and it doesn't mention that the range = codomain?)

So I guess there are only 4 sentences that describe an onto function here

 

[castor28]

Point 1 was correct.

$f(A)$ is the range, $B$ is the co-domain. $f$ is onto if the range equals the co-domain, i.e., if $f(A) = B$.

 

[lemonthree]

Point 1 was correct.

$f(A)$ is the range, $B$ is the co-domain. $f$ is onto if the range equals the co-domain, i.e., if $f(A) = B$.

I see, thank you for your explanation.