Let f, g: ℝ → ℝ be defined by f(x) = x² and g(x) = x. What is the equalizer of f and g in Set?
AThe empty set, because f and g never agree
BThe set {0, 1} with the inclusion into ℝ, because those are the points where x² = x
CThe set ℝ itself, because any function from any set factors through f and g
DThe set {1} only, because 1 is the multiplicative identity for both f and g
The equalizer in Set is {x ∈ A | f(x) = g(x)} = {x ∈ ℝ | x² = x} = {x | x(x−1) = 0} = {0, 1}, together with the inclusion map e: {0, 1} → ℝ. The universal property: any function h: C → ℝ satisfying f∘h = g∘h must only pick points where x² = x, so it lands in {0, 1} and factors uniquely through the inclusion. The equalizer is the 'largest subobject of ℝ where f and g agree.'
Question 2 Multiple Choice
In the category Ab (abelian groups), the kernel of a homomorphism φ: G → H is a special case of an equalizer. What is the second parallel morphism that φ is being equalized against?
AThe identity morphism id: G → H
BThe zero morphism 0: G → H that sends every element of G to the identity element of H
CThe inverse morphism φ⁻¹: H → G
DThe projection morphism π: G → G/ker(φ)
The kernel ker(φ) = {g ∈ G | φ(g) = e_H} is exactly the set where φ agrees with the zero morphism (which sends every g to e_H). So ker(φ) is the equalizer of φ and 0: G → H. The inclusion k: ker(φ) → G satisfies φ∘k = 0∘k (both send every element to e_H), and it is universal: any h: C → G with φ∘h = 0 factors uniquely through k. This shows that 'kernel' is not a standalone concept but an instance of the universal equalizer construction.
Question 3 True / False
The coequalizer of f, g: A → B in Set is a quotient of A that identifies elements where f(a) = g(a).
TTrue
FFalse
Answer: False
The coequalizer is a quotient of *B*, not A. Given f, g: A → B, the coequalizer Q = B/~ where ~ is generated by f(a) ~ g(a) for all a ∈ A. The coequalizer map goes q: B → Q. You are identifying elements *in the codomain B* that are related via f and g, not elements of the domain A. This is the dual of the equalizer: the equalizer restricts the domain to where f and g agree; the coequalizer collapses the codomain to force f and g to become equal after post-composition.
Question 4 True / False
An equalizer morphism e: E → A is generally a monomorphism (injective on hom-sets) in any category.
TTrue
FFalse
Answer: False
Equalizers are monomorphisms in Set and in most algebraic categories (groups, rings, modules), but not in every category. The statement is true in categories that are sufficiently 'concrete' (well-powered, with enough structure). However, in an arbitrary category, an equalizer need not be monic. The universal property guarantees uniqueness of factorization but does not, by itself, force the equalizer morphism to be a monomorphism in general. This is a case where Set-based intuition must be checked against the categorical definition.
Question 5 Short Answer
Explain in your own words the conceptual difference between an equalizer and a coequalizer — what problem does each solve, and how do the answers differ structurally?
Think about your answer, then reveal below.
Model answer: An equalizer asks: 'Where do f and g already agree?' It finds the largest sub-object of the domain A from which both morphisms give identical outputs, together with the inclusion into A. Structurally, it is a restriction — a subobject. A coequalizer asks: 'How do we force f and g to become equal?' It builds the smallest quotient of the codomain B in which f(a) and g(a) are identified for every a ∈ A, together with the quotient map from B. Structurally, it is a collapse — a quotient object. They are dual: the equalizer operates on the domain side by inclusion; the coequalizer operates on the codomain side by identification.
This duality is the categorical heart of the constructions. Every concept about equalizers (universal property, monomorphism property in nice categories) has a dual version for coequalizers (universal property, epimorphism property). In abelian categories, equalizers give kernels (subobjects) and coequalizers give cokernels (quotient objects), and the interplay between these is what makes exact sequences and homological algebra possible.