Let M be the free monoid on {a, b}, and let N be the monoid of non-negative integers under addition. We define f(a) = 3 and f(b) = 5. By the universal property, what is the unique monoid homomorphism f̄: M → N extending f?
Af̄ is not uniquely determined; we have freedom to assign values to compound words like 'ab' however we like
Bf̄(ab) = 8, f̄(ba) = 8, f̄(aab) = 11, and generally f̄ maps each word to the sum of the integer values of its letters
Cf̄(ab) = 15 because concatenation in M corresponds to multiplication in N
DThe universal property does not apply here because N and M have different cardinalities
The homomorphism law forces the extension: f̄ must satisfy f̄(wv) = f̄(w) + f̄(v) for all words w, v (since the operation in M is concatenation and in N is addition). This uniquely determines f̄ on every word: f̄(ab) = f̄(a) + f̄(b) = 3 + 5 = 8, f̄(aab) = 3 + 3 + 5 = 11. There is no choice — the homomorphism condition and the values on generators completely determine the entire function. Option A is the misconception: once you specify generators, you have no freedom left. This is precisely what the universal property asserts.
Question 2 Multiple Choice
Which statement best captures what makes the free group on generators S 'free'?
AThe free group has only finitely many elements, one for each generator and its inverse
BThe free group imposes no algebraic relations beyond those required by the group axioms; every other group generated by the same number of generators is a quotient of it
CThe free group is the trivial group containing only the identity, since no relations force any non-trivial elements
DThe free group exists in every algebraic category because the forgetful functor always has a left adjoint
The free group on S contains all reduced words in the generators and their formal inverses, with no extra equalities imposed — ab ≠ ba, a ≠ a⁻¹, and so on. Any group generated by a set in bijection with S is a quotient of the free group: you obtain it by imposing additional relations (e.g., ab = ba gives the free abelian group; a² = e gives a quotient with specific structure). The free object is the 'most general' in the sense that it has the fewest constraints, and all others are obtained by imposing more. Option D is wrong: not every category has free objects; the forgetful functor needs a left adjoint, which is not guaranteed.
Question 3 True / False
The free monoid on a two-element set {a, b} contains infinitely many distinct elements.
TTrue
FFalse
Answer: True
The free monoid consists of all finite words (sequences) in {a, b}: the empty word ε, then a, b, aa, ab, ba, bb, aaa, aab, aba, ..., and so on for every finite string. Since there is no upper bound on word length, there are infinitely many distinct elements. Two words are equal only if they are identical character by character — no extra relations collapse distinct words. This is the 'no extra relations beyond the axioms' property of free objects: the monoid axioms do not force any word to equal any other word except through the identity laws (εw = w = wε).
Question 4 True / False
Given a function f: S → U(M) assigning each generator to an element of a monoid M, there may be multiple monoid homomorphisms from the free monoid F(S) → M extending f, and we are free to choose among them.
TTrue
FFalse
Answer: False
The universal property of the free monoid guarantees exactly ONE extension — uniqueness is the whole point. Once you specify where each generator goes, the homomorphism law uniquely determines the value on every word: f̄(ab) must equal f̄(a)·f̄(b), f̄(aab) must equal f̄(a)·f̄(a)·f̄(b), and so on. There is no choice. This uniqueness is what distinguishes the free object from other objects — the bijection Hom_C(F(S), A) ≅ Hom_Set(S, U(A)) is natural precisely because each function from generators extends to exactly one homomorphism.
Question 5 Short Answer
State the universal property of the free monoid on a set S in your own words. Why does this property mean that defining a monoid homomorphism out of the free monoid is equivalent to choosing where the generators go?
Think about your answer, then reveal below.
Model answer: The universal property says: for any monoid M and any function f: S → M, there exists a unique monoid homomorphism f̄: Free(S) → M such that f̄(s) = f(s) for all s ∈ S. Defining a homomorphism out of Free(S) requires specifying f̄ on all words; but the homomorphism law forces f̄(s₁s₂...sₙ) = f(s₁)·f(s₂)·...·f(sₙ), so the images of the generators completely and uniquely determine f̄. There is nothing more to specify.
This bijection Hom_Monoid(Free(S), M) ≅ Hom_Set(S, U(M)) is the adjunction hom-set bijection, expressing that the free construction is left adjoint to the forgetful functor. In practice it means: to construct a monoid homomorphism out of a free monoid, you just assign each generator an image in the target monoid — the rest is forced. This is why free objects are so useful in algebra: presentations of groups, rings, and modules as 'generators and relations' always start with a free object (the generators with no relations) and then quotient by the desired relations.