Is this proof of the Pigeonhole Principle acceptable?

[Agent M27]

Homework Statement

I have posted my proof of the Pigeonhole Principle.

Homework Equations

The Attempt at a Solution

Basically I am curious if this is an acceptable proof of the Pigeonhole Principle? I ask because both my professor and our textbook complete it differently, both of which make use of the restriction of a function. I choose this form for two simple reasons, I am having a hard time understanding why they are employing the restriction of a function and why they are employing the function g, but also this seems a lot simpler. I tend to make these things more difficult than they ought to be, but in mathematics nothing is arbitrary except the objects we place in sets. I have included both my version and the version my book used. My professor gave a proof similar to the one in the book. Thanks in advance. The first screenshot is my proof.

Joe

 

[lippyka]

Proof:Let A and B be two finite sets with |A| = m and |B| = n. Assume that there is a function f : A -> B such that each element of A is associated with an element of B. We will prove that if m > n, then there must exist two elements in A that are associated with the same element in B.Suppose by way of contradiction that no two elements of A are associated with the same element of B. Then, since |A| = m and |B| = n, each element in B must be associated with at least one element in A.But this implies that |A| ≤ |B|, which contradicts the assumption that m > n. Therefore, two elements of A must be associated with the same element of B. Q.E.D.Book's version:Let A and B be two finite sets with |A| = m and |B| = n. Assume that there is a function g : A -> B. We will prove that if m > n, then there must exist two elements in A that are associated with the same element in B.Suppose by way of contradiction that no two elements of A are associated with the same element in B. Then, since |A| = m and |B| = n, each element in B must be associated with exactly one element in A.But this implies that |A| = |B|, which contradicts the assumption that m > n. Therefore, two elements of A must be associated with the same element of B. Q.E.D.