Exercise 1: Classifying f(1)=p, f(2)=q, f(3)=p, f(4)=q — Possible Solution
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CHECKING INJECTIVE
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Injective requires that different inputs always produce different
outputs. Checking: f(1) = p and f(3) = p - two different inputs (1
and 3) produce the SAME output (p). This single collision is enough
to break injectivity. f is NOT injective.

CHECKING SURJECTIVE
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Surjective requires that every element of the codomain B = {p, q}
actually gets hit by at least one input. Checking: p is hit (by both
1 and 3). q is hit (by both 2 and 4). Both elements of B are
accounted for, so f IS surjective.

CHECKING BIJECTIVE
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Bijective requires both injective and surjective at once. Since f
already fails injective, it cannot be bijective, regardless of the
surjective result.

SUMMARY
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f is surjective, but not injective, and therefore not bijective.

WHY THIS RESULT ALSO FOLLOWS FROM THIS CHAPTER'S OWN CARDINALITY
REASONING
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Per this chapter, "if |A| > |B|, no injective function from A to B
can exist at all." Here |A| = 4 (the set {1,2,3,4}) and |B| = 2 (the
set {p,q}) - since 4 > 2, no function from A to B could ever be
injective in the first place, regardless of how it's defined. The
specific collision found above (f(1)=f(3)=p) is simply one concrete
instance of a mismatch that was already guaranteed to exist by the
size difference alone.

WHY THIS WORKS AS AN ANSWER
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It checks injectivity and surjectivity directly against the given
function's own mapping, correctly concludes bijectivity fails because
injectivity already failed, and connects the result back to this
chapter's own cardinality principle to show the outcome wasn't
arbitrary - it was guaranteed in advance by the relative sizes of the
two sets.