All but two of the following statements are correct ways to express the fact that a function $ f $ is onto. Find the two that are incorrect.

a. $ f $ is onto $\Leftrightarrow$ every element in its co-domain is the image of some element in its domain.

b. $ f $ is onto $\Leftrightarrow$ every element in its domain has a corresponding image in its co-domain.

c. $ f $ is onto $\Leftrightarrow$ $\forall y \in Y, \exists x \in X$ such that $ f(x) = y $.

d. $ f $ is onto $\Leftrightarrow$ $\forall x \in X, \exists y \in Y$ such that $ f(x) = y $.

e. $ f $ is onto $\Leftrightarrow$ the range of $ f $ is the same as the co-domain of $ f $.

The two incorrect statements are: c. f is onto ⇔ ∀y ∈ Y, 3x ∈ X such that f(x)= y. (The correct quantifier is ∃, not ∀.). d. f is onto ⇔ ∀x ∈ X, 3y ∈ Y such that f(x)= y. (The correct quantifier is ∃, not ∀.)

The correct statements are: a. function f is onto ⇔ every element in its co-domain is the image of some element in its domain. b. f is onto ⇔ every element in its domain has a corresponding image in its co-domain. e. f is onto ⇔ the range of f is the same as the co-domain of f.

To further explain the correct statements:

a. This statement is a direct definition of an onto function. It means that for every element y in the co-domain of f, there exists some element x in the domain of f such that f(x) = y.

b. This statement is equivalent to statement a and is also a direct definition of an onto function. It means that for every element x in the domain of f, there exists some element y in the co-domain of f such that f(x) = y.

e. This statement is also equivalent to statements a and b. The range of f is the set of all possible outputs that f can produce, while the co-domain of f is the set of all possible outputs that f could produce. Therefore, if the range of f is equal to its co-domain, then every possible output of f is actually produced by f, which means that f is onto.

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