advanced34 min studyLesson 535 in Mathematics

Products, coproducts, and equalizers

Build finite universal constructions and verify their mapping directions explicitly.

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What you’ll understand

  • Verify product and coproduct mapping properties without relying only on cardinality.
  • Construct an equalizer and its unique factorization map.
  • Distinguish a tagged coproduct from an overlapping set union.

Products combine independently specified outputs, coproducts combine independently specified inputs, and equalizers impose agreement between two arrows. Their concrete set constructions are familiar, but their universal properties explain which parts survive in other categories. The direction of the test map is the central organizing choice. We will calculate an equalizer, count all compatible maps into it, and then compare product and coproduct counts where an ordinary union would give the wrong answer.

Products are universal joint receivers

A product of AA and BB consists of an object PP and projections to the two factors. Given any test object TT with maps to both factors, there must be a unique map from TT to PP producing those two maps after projection. In Set, ordered pairs provide this map point by point. The universal property characterizes the product with its projections, not an arbitrary object having the same number of elements.

For each TT, the property is a bijection from C(T,P)\mathcal C(T,P) to a product of Hom sets, natural in the test object. Empty products are terminal objects because the empty family of maps has exactly one compatible realization. A category need not possess products at all. When two products do exist, applying their mapping properties to each other produces inverse comparison maps, uniquely determined by their compatibility with both projections.

C(T,A×B)≅C(T,A)×C(T,B).\mathcal C(T,A\times B)\cong\mathcal C(T,A)\times\mathcal C(T,B).
A product receives a single map encoding a pair of maps from the same test object.↔ Scroll wide equations horizontally.

Coproducts reverse the mapping problem

A coproduct has injections from AA and BB and is universal for extending a pair of maps out of these objects to one map out of the coproduct. In Set, a tagged disjoint union keeps an element of the first summand distinct from an equal-looking element of the second. The tags are needed because the two original maps may prescribe different outputs at that shared-looking element. An overlapping ordinary union would accidentally force those outputs to agree.

The mapping property now tests maps into an arbitrary target TT. This reversal distinguishes a coproduct from a product even when a particular category happens to realize both with the same underlying object. Empty coproducts are initial objects. A map from a coproduct is determined by its restrictions to summands; a map into a product is determined by its projections. Neither statement authorizes interchanging the two directions in an unfamiliar category.

C(A⊔B,T)≅C(A,T)×C(B,T).\mathcal C(A\sqcup B,T)\cong\mathcal C(A,T)\times\mathcal C(B,T).
The test object is now the target, and the universal map leaves the coproduct.↔ Scroll wide equations horizontally.

Equalizers impose an equation universally

For parallel arrows f,g:X→Yf,g:X\to Y, an equalizer is an arrow e:E→Xe:E\to X with f∘e=g∘ef\circ e=g\circ e. It must satisfy more than this equation: whenever h:T→Xh:T\to X also equalizes the pair, there is exactly one u:T→Eu:T\to E with e∘u=he\circ u=h. In Set, take the subset of points at which the functions agree and use its inclusion. Every equalizing function lands in that subset, so restricting its codomain supplies the factorization.

This construction is a subobject of the source, while the dual coequalizer is a quotient of the target identifying outputs that must agree. Confusing them reverses the equation being imposed. In a general category an equalizer arrow is monic: if two maps become equal after composition with it, they are two factorizations of the same equalizing map and hence coincide. The familiar injectivity of a subset inclusion is therefore an instance of a universal argument.

E={x∈X:f(x)=g(x)},e∘u=huniquely whenever f∘h=g∘h.E=\{x\in X:f(x)=g(x)\},\qquad e\circ u=h\quad\text{uniquely whenever }f\circ h=g\circ h.
The subset construction in Set satisfies both the equation and the unique factorization requirement.↔ Scroll wide equations horizontally.

An equalizer with eight compatible test maps

Let X={a,b,c,d}X=\{a,b,c,d\} and Y={0,1,2}Y=\{0,1,2\}. In that source order, set f=(0,1,1,2)f=(0,1,1,2) and g=(0,2,1,0)g=(0,2,1,0). Construct the equalizer and factor the function h:{p,q,r}→Xh:\{p,q,r\}\to X with values (a,c,a)(a,c,a). Count every possible equalizing function from this test set.

  1. Compare the functions at all four source points. They agree at a, both with value zero, and at c, both with value one. They disagree at b and d. Thus the equalizer is E={a,c}E=\{a,c\} with its inclusion into XX.

  2. The given function lands entirely in EE. Its factorization sends p to a, q to c, and r to a, now regarded as elements of EE. The inclusion cannot hide a second value, so this factorization is unique.

  3. Every equalizing function from the three-point test set must choose a or c independently at each point. Conversely, each such choice equalizes the functions. There are therefore eight compatible maps, each corresponding to exactly one map into the equalizer.

E={a,c},∣Set({p,q,r},E)∣=23=8.E=\{a,c\},\qquad |\mathbf{Set}(\{p,q,r\},E)|=2^3=8.
The count follows from the actual factorization bijection, not merely from a guessed subset size.↔ Scroll wide equations horizontally.
The result

The equalizer has two elements, the stated factorization is unique, and the three-point probe has eight equalizing maps.

Explore the model

Interactive lab

Interactive lab · modular

See a finite equalizer as an orbit

Pause and Reset, then set modulus to 6 and additive step to 3. Predict the orbit {0,3} and check its two-step period. On the six residues, compare the functions r↦2r modulo 6 and r↦0; their equalizer is exactly {0,3}. The instrument enumerates that orbit, while the unique factorization of every equalizing function must be checked separately. Before a fresh comparison use Pause, Reset, and set both controls, because a control change retains elapsed time.

See a finite equalizer as an orbitAddition modulo m starts at zero: rₙ = na mod m. Playback advances two additions per second. The orbit has period m/gcd(m,a), including period one for a = 0. This is not a cryptographic security test or a primality algorithm. Current readings: Residue r0: 0; Orbit period · m/gcd(m,a): 12; Distinct residues visited: 1 of 12. All parameters are adjustable with the labeled controls below.Repeated addition modulo 12: the current residue is 0, and the orbit period is 12.01234567891011+ 5mod 12An additive orbitn = 0rₙ = 0Period = 12Cycle starts at 0:0 → 5 → 10 → 3 →8 → 1 → 6 → 11 →4 → 9 → 2 → 7 → 0Exact residues · two additions per second
Additive orbitCurrent residue / next edge
Residue r0
0
Orbit period · m/gcd(m,a)
12
Distinct residues visited
1 of 12
t = 0 s

Addition modulo m starts at zero: rₙ = na mod m. Playback advances two additions per second. The orbit has period m/gcd(m,a), including period one for a = 0. This is not a cryptographic security test or a primality algorithm. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forAn orbit may happen to realize a particular equalizer subset, but the universal mapping property is additional structure.

Check your understanding

For a coproduct A⊔B, which pair of maps determines a unique induced map?

Try it yourself

In Set, let A={0,1}A=\{0,1\} and B={1,2,3}B=\{1,2,3\}. Find the sizes of the product and coproduct. For a two-element set TT, count maps T→A×BT\to A\times B and maps A⊔B→TA\sqcup B\to T, checking both mapping properties.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The shared symbol one does not identify the two tagged summand elements. Count functions from a finite domain by independent choices of output.

Compare with the solution

The product has six elements and the tagged coproduct has five; the ordinary union has only four and is not the coproduct with these inclusions. There are 62=366^2=36 maps from TT into the product, matching 22⋅32=4⋅92^2\cdot3^2=4\cdot9. There are 25=322^5=32 maps from the coproduct into TT, matching 22⋅23=4⋅82^2\cdot2^3=4\cdot8. The two formulas test opposite mapping directions.

Take it with you

  • Maps into a product encode pairs of maps into its factors.
  • Maps out of a coproduct encode independent maps out of its summands.
  • An equalizer universally factors maps satisfying a parallel-arrow equation.

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