All but two of the following statements are correct ways to express the fact that a function $ f $ is onto. Select 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 \text{ such that } f(x) = y $.

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

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

Let's go through each statement and explain why they are incorrect:

1. The statement "f is onto ⇔ ∀y∈Y,∃x∈X such that f(x)=y" is incorrect because it mistakenly implies that for every y in the co-domain, there exists exactly one x in the domain such that f(x) equals y.

However, in an onto function, it is possible for multiple elements in the domain to map to the same element in the co-domain.

So, the correct statement should be "f is onto ⇔ for every y in the co-domain, there exists at least one x in the domain such that f(x) equals y."

2. The statement "f is onto ⇔ the range of f is the same as the co-domain of f" is incorrect because it wrongly suggests that the range and the co-domain of a function are always equal for an onto function.

However, the range of a function is the set of all possible output values, while the co-domain is the set that contains all the possible output values, including those that may not be mapped from the domain.

In an onto function, the range will be equal to the co-domain, but the converse is not always true. So, the correct statement should be "f is onto ⇔ the range of f is a subset of the co-domain of f."

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