advanced34 min studyLesson 539 in Mathematics

Equivalences, localizations, and inverting morphisms

Compare categorical presentations and enforce inversion by a universal mapping property.

On this page

What you’ll understand

  • Distinguish equivalence of categories from an isomorphism of categories.
  • State localization through the functors and transformations that it universally extends.
  • Compute a finite localization and distinguish it from a finite cyclic representation.

Equivalent categories can have different numbers of objects while presenting the same structure up to isomorphism. Localization changes which arrows are invertible and is characterized by what other functors it allows. Both ideas avoid insisting on equality of concrete presentations, but they solve different problems. We will invert the single nonidentity arrow of a finite category, then contrast that universal construction with an ordinary representation of an infinite group in a finite cyclic group.

Equivalence allows isomorphic redundancy

A functor F:C→DF:\mathcal C\to\mathcal D is an equivalence when there is a functor GG in the other direction with natural isomorphisms GF≅IdCGF\cong\mathrm{Id}_{\mathcal C} and FG≅IdDFG\cong\mathrm{Id}_{\mathcal D}. An isomorphism of categories would require strict inverse equalities instead. Equivalent categories can therefore differ by redundant isomorphic objects, while an isomorphism of categories gives a bijection on the actual object sets.

A full and faithful functor that is essentially surjective is an equivalence, with the usual choice of representatives and comparison isomorphisms; for small categories ordinary set-theoretic choice suffices. Essential surjectivity means every target object is isomorphic to an image object, rather than literally equal to one. Full faithfulness supplies the unique lifted arrows needed to build a quasi-inverse after representatives are chosen. Merely being surjective on objects would not recover missing arrows or distinguish identified arrows.

GF≅IdC,FG≅IdD.GF\cong\mathrm{Id}_{\mathcal C},\qquad FG\cong\mathrm{Id}_{\mathcal D}.
The comparisons are natural isomorphisms rather than required equalities of functors.↔ Scroll wide equations horizontally.

Localization universally permits specified inverses

Let C\mathcal C be small and W a set of its arrows. A localization Q:C→C[W−1]Q:\mathcal C\to\mathcal C[W^{-1}] sends every arrow in W to an isomorphism. For each target category D, precomposition with Q is fully faithful, and its essential image consists of the W-inverting functors. Equivalently, it is an equivalence onto the full subcategory of those functors. The statement includes natural transformations as well as the existence of factorizations, not just an object-level assignment of functors.

One construction uses finite composable zigzags, permitting the selected arrows to be traversed backward, and quotients by composition, identity, and inverse relations. Smallness makes the relevant paths and their quotient into sets. In this explicit presentation, extensions can be made strictly unique by evaluating backward arrows as inverses. Arbitrary localizations express that uniqueness up to the appropriate natural isomorphism. Reducing every zigzag to a single fraction-shaped roof requires extra calculus-of-fractions hypotheses and is not automatic for an arbitrary W.

Q∗:Fun⁡(C[W−1],D)≃Fun⁡W(C,D).Q^*:\operatorname{Fun}(\mathcal C[W^{-1}],\mathcal D)\simeq\operatorname{Fun}_W(\mathcal C,\mathcal D).
The right side is the full subcategory of functors carrying every selected arrow to an isomorphism.↔ Scroll wide equations horizontally.

A finite image is not a universal localization

The one-object category with endomorphism monoid N\mathbb N under addition is generated by one arrow. Inverting that generator gives the one-object category with group Z\mathbb Z: a target functor is determined by an invertible endomorphism a, and the extension sends an integer n to its positive or negative power. No finite-order relation is imposed unless it already follows from the original relations. This is why formal inversion produces integers rather than residues modulo a chosen modulus.

A homomorphism from the integers to a finite cyclic group is a useful W-inverting representation, but it identifies distinct integers and cannot replace the universal construction. Similarly, making some arrows invertible can identify other arrows through the imposed relations; localization need not be faithful. Its guarantee concerns all W-inverting functors out of the original category. This mapping property determines the resulting category up to equivalence compatible with Q, without promising that its Hom sets retain their old sizes.

N[1−1]≅Z,F‾(n)=an(n∈Z).\mathbb N[1^{-1}]\cong\mathbb Z,\qquad \overline F(n)=a^n\quad(n\in\mathbb Z).
Here 1 denotes the additive generator, and negative exponents refer to the inverse of its image.↔ Scroll wide equations horizontally.

Invert one arrow in a two-object category

Let C\mathcal C have two objects zero and one, their identity arrows, and one additional arrow f:0→1f:0\to1. Localize at f. List all arrows of the resulting category and show that it is equivalent, but not isomorphic, to the terminal category.

  1. Adjoin g:1→0g:1\to0 with gf=id0gf=\mathrm{id}_0 and fg=id1fg=\mathrm{id}_1. Every longer path reduces using these relations, so the only arrows are the two identities, f, and g. There is exactly one arrow of each source-target type.

  2. A functor out of the original category sends f to some arrow h. It extends precisely when h is invertible, and the image of g must then be its unique inverse. The relations are preserved because the two inverse identities hold in the target.

  3. The unique functor to the terminal category has a quasi-inverse selecting zero. The composite on the two-object category is naturally isomorphic to its identity, with components id0\mathrm{id}_0 and f. There cannot be a strict category isomorphism to a one-object category, because the object sets have different sizes.

C[f−1](i,j)={∗}(i,j∈{0,1}),∣Mor⁡∣=4.\mathcal C[f^{-1}](i,j)=\{*\}\quad(i,j\in\{0,1\}),\qquad |\operatorname{Mor}|=4.
All paths reduce to the unique arrow with their given endpoints.↔ Scroll wide equations horizontally.
The result

The localization is the four-arrow groupoid with two uniquely isomorphic objects. It is equivalent to a one-object terminal category while retaining two objects in this presentation.

Explore the model

Interactive lab

Interactive lab · modular

An inverse in a finite representation

Pause and Reset, then set modulus to 6 and additive step to 2. One Step selects residue 2. For an independent run, Pause and Reset, then set modulus to 6 and step to 4; one Step selects 4. Their sum is zero modulo 6, so these endomorphisms are inverse under additive composition. The image generated by 2 has only three residues; it is a finite representation of the integer generator, not the universal localization itself. Changing a control without Reset retains elapsed time.

An inverse in a finite representationAddition 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 invertible image satisfies the localization requirement but may impose extra finite-order relations.

Check your understanding

After inverting the generator of the one-object monoid category (N,+), why is Z universal rather than Z/6Z?

Try it yourself

In the localization of the one-object monoid (N,+)(\mathbb N,+) at its generator, simplify the zigzag value 3−5+13-5+1. If a target functor sends the generator to an invertible endomorphism a, determine the image of this zigzag and explain what a reduction modulo six would lose.

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

Give me a hint

Compose by addition of signed exponents. Then test the proposed finite quotient against an invertible element of infinite order.

Compare with the solution

The signed exponent is minus one, so the universal extension sends the zigzag to a−1a^{-1}. Modulo six this exponent has the same residue as five, but those exponents need not have equal images when a has infinite order. A target such as the additive integer group distinguishes them. Therefore reduction modulo six gives one representation, not the universal inversion property.

Take it with you

  • Equivalence permits redundant isomorphic objects, whereas category isomorphism requires strict inverses.
  • Localization extends exactly the functors that invert the selected arrows, including their natural transformations.
  • A finite representation can impose additional relations absent from a universal localization.

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