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I am a set theorist in my orientation, and while I did take a few courses that brushed upon categorical and algebraic constructions, one has always eluded me.

The inverse limit. I tried to ask one of the guys in my office, and despite a very shady explanation he ended up muttering that "you usually take an already known construction."

The Wikipedia article presents two approaches, the algebraic and the categorical. While the categorical is extremely vague for me, the algebraic one is too general and the intuition remains hidden underneath the text in a place I cannot find it.

Since I am not too familiar with categories, the explanation most people would try to give me which is categorical in nature seems to confuse me - as I keep asking this question over and over every now and then.

Could anyone explain to me in non-categorical terms what is the idea behind an inverse limit? (I am roughly familiar with its friend "direct limit", if that helps)

(While editing, I can say that the answers given so far are very interesting, and I have read them thoroughly, although I need to give it quite some thinking before I can comment on all of them right now.)

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I was reluctant to tag this under category related tags, as I'm really asking to avoid categories. If anyone thinks of something better to retag it with, be my guest. –  Asaf Karagila May 11 '11 at 18:11
If you're going to ask "Please tell me about X in basic terms", and there is a wikipedia article on X, I think it would be a good idea to read and make reference to the article, as what it says is going to be along the lines of how many will try to answer. (In this case you have gotten some very creative answers. If I were to answer it would be much more along the lines of wikipedia, in part because that's the kind of explanation I would have found helpful as a beginner in this area.) –  Pete L. Clark May 11 '11 at 19:53
@Pete: I'm very glad to see you back on the site. I thought that I did mention that I read the wikipedia page and it wasn't very clear to me in the sense that the intuition was hidden and the categorical abstract definition was the center of the text. –  Asaf Karagila May 11 '11 at 20:06
This looks like a duplicate of math.stackexchange.com/questions/19838/… to me –  Alex B. May 12 '11 at 5:24
@Alex: It looks to me as though this question is a generalization of that question, and the answers here are slightly more general than the ones in the question you linked. I do agree that the topics are closely related - I did search for "inverse limit" but not for "projective limit" prior to posting this question. (All this regardless as being the OP, and as unbiased as I can possibly be on the matter.) –  Asaf Karagila May 12 '11 at 6:48

6 Answers 6

up vote 113 down vote accepted

I like George Bergman's explanation (beginning in section 7.4 of his Invitation to General Algebra and Universal Constructions.

We start with a motivating example.

Suppose you are interested in solving $x^2=-1$ in $\mathbb{Z}$. Of course, there are no solutions, but let's ignore that annoying reality for a moment.

The equation has a solution in the ring $\mathbb{Z}_5$ (in fact, two: both $2$ and $3$, which are the same up to sign). So we want to find a solution to $x^2=-1$ in $\mathbb{Z}$ which satisfies $x\equiv 2 \pmod{5}$.

An integer that is congruent to $2$ modulo $5$ is of the form $5y+2$, so we can rewrite our original equation as $(5y+2)^2 = -1$, and expand to get $25y^2 + 20y = -5$.

That means $20y\equiv -5\pmod{25}$, or $4y\equiv -1\pmod{5}$, which has the unique solution $y\equiv 1\pmod{5}$. Substituting back we determine $x$ modulo $25$: $$x = 5y+2 \equiv 5\cdot 1 + 2 = 7 \pmod{25}.$$

Continue this way: putting $x=25z+7$ into $x^2=-1$ we conclude $z\equiv 2 \pmod{5}$, so $x\equiv 57\pmod{125}$.

Using Hensel's Lemma, we can continue this indefinitely. What we deduce is that there is a sequence of residues, $$x_1\in\mathbb{Z}_5,\quad x_2\in\mathbb{Z}_{25},\quad \ldots, x_{i}\in\mathbb{Z}_{5^i},\ldots$$ each of which satisfies $x^2=-1$ in the appropriate ring, and which are "consistent", in the sense that each $x_{i+1}$ is a lifting of $x_i$ under the natural homomorphisms $$\cdots \stackrel{f_{i+1}}{\longrightarrow} \mathbb{Z}_{5^{i+1}} \stackrel{f_i}{\longrightarrow} \mathbb{Z}_{5^i} \stackrel{f_{i-1}}{\longrightarrow}\cdots\stackrel{f_2}{\longrightarrow} \mathbb{Z}_{5^2}\stackrel{f_1}{\longrightarrow} \mathbb{Z}_5.$$

Take the set of all strings $(\ldots,x_i,\ldots,x_2,x_1)$ such that $x_i\in\mathbb{Z}_{5^i}$ and $f_i(x_{i+1}) = x_i$, $i=1,2,\ldots$. This is a ring under componentwise operations. What we did above shows that in this ring, you do have a square root of $-1$.

Added. Bergman here inserts the quote, "If the fool will persist in his folly, he will become wise." We obtained the sequence by stubbornly looking for a solution to an equation that has no solution, by looking at putative approximations, first modulo 5, then modulo 25, then modulo 125, etc. We foolishly kept going even though there was no solution to be found. In the end, we get a "full description" of what that object must look like; since we don't have a ready-made object that satisfies this condition, then we simply take this "full description" and use that description as if it were an object itself. By insisting in our folly of looking for a solution, we have become wise by introducing an entirely new object that is a solution.

This is much along the lines of taking a Cauchy sequence of rationals, which "describes" limit point, and using the entire Cauchy sequence to represent this limit point, even if that limit point does not exist in our original set.

This ring is the $5$-adic integers; since an integer is completely determined by its remainders modulo the powers of $5$, this ring contains an isomorphic copy of $\mathbb{Z}$.

Essentially, we are taking successive approximations to a putative answer to the original equation, by first solving it modulo $5$, then solving it modulo $25$ in a way that is consistent with our solution modulo $5$; then solving it modulo $125$ in a way that is consistent with out solution modulo $25$, etc.

The ring of $5$-adic integers projects onto each $\mathbb{Z}_{5^i}$ via the projections; because the elements of the $5$-adic integers are consistent sequences, these projections commute with our original maps $f_i$. So the projections are compatible with the $f_i$ in the sense that for all $i$, $f_i\circ\pi_{i+1} = \pi_{i}$, where $\pi_k$ is the projection onto the $k$th coordinate from he $5$-adics.

Moreover, the ring of $5$-adic integers is universal for these property: given any ring $R$ with homomorphisms $r_i\colon R\to\mathbb{Z}_{5^i}$ such that $f_i\circ r_{i+1} = r_i$, for any $a\in R$ the tuple of images $(\ldots, r_i(a),\ldots, r_2(a),r_1(a))$ defines an element in the $5$-adics. The $5$-adics are the inverse limit of the system of maps $$\cdots\stackrel{f_{i+1}}{\longrightarrow}\mathbb{Z}_{5^{i+1}}\stackrel{f_i}{\longrightarrow}\mathbb{Z}_{5^i}\stackrel{f_{i-1}}{\longrightarrow}\cdots\stackrel{f_2}{\longrightarrow}\mathbb{Z}_{5^2}\stackrel{f_1}{\longrightarrow}\mathbb{Z}_5.$$

So the elements of the inverse limit are "consistent sequences" of partial approximations, and the inverse limit is a way of taking all these "partial approximations" and combine them into a "target object."

More generally, if you have a system of, say, rings, $\{R_i\}$, indexed by an directed set $(I,\leq)$ (so that for all $i,j\in I$ there exists $k\in I$ such that $i,j\leq k$, and a system of maps $f_{rs}\colon R_s\to R_r$ whenever $r\leq s$ which are "consistent" (if $r\leq s\leq t$, then $f_{st}\circ f_{rs} = f_{rt}$), and let's assume that the $f_{rs}$ are surjective, as they were in the example of the $5$-adics. Then you can think of the $R_i$ as being "successive approximations" (with a higher indexed $R_i$ as being a "finer" or "better" approximation than the lower indexed one). The directedness of the index set guarantees that given any two approximations, even if they are not directly comparable to one another, you can combine them into an approximation which is finer (better) than each of them (if $i,j$ are incomparable, then find a $k$ with $i,j\leq k$). The inverse limit is a way to combine all of these approximations into an object in a consistent manner.

If you imagine your maps as going right to left, you have a branching tree that is getting "thinner" as you move left, and the inverse limit is the combination of all branches occurring "at infinity".

Added. The example of the $p$-adic integers may be a bit misleading because our directed set is totally ordered and all maps are surjective. In the more general case, you can think of every chain in the directed set as a "line of approximation"; the directed property ensures that any finite number of "lines of approximation" will meet in "finite time", but you may need to go all the way to "infinity" to really put all the lines of approximation together. The inverse limit takes care of this.

If the directed set has no maximal elements, but the structure maps are not surjective, it turns out that no element that is not in the image will matter; essentially, that element never shows up in a net of "successive approximations", so it never forms part of a "consistent system of approximations" (which is what the elements of the inverse limit are).

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@Asaf: For one thing, every nonzero element of the inverse limit has infinite additive order, but the direct limit contains elements of order $5$ (the image of the generator of $\mathbb{Z}_5$ in the direct limit has order $5$); in fact, the direct limit is torsion, the inverse limit is torsionfree. In the direct limit you are forcing global conformity/consistency (by identifying elements that are eventually mapped to the same thing, even if they are different objects to being with), whereas in the inverse limit you are requiring local conformity. –  Arturo Magidin May 11 '11 at 19:22
@Asaf: he direct limit is the Prufer group $\mathbb{Z}_{5^{\infty}}$. The direct limit is a quotient of the direct sum, the inverse limit is a subgroup of the direct product (there's that categorical duality you didn't want...) The examples are possibly misleading in that they have very special structure maps and index sets... –  Arturo Magidin May 11 '11 at 19:46
@Asaf: In the direct limit, it's as if you are "pulling away" and seeing a larger and larger vista, or more generally, you are seeing bigger and bigger pictures, and you glue them together by forcing the overlaps to agree. In the inverse limits, you are "zooming in" and looking at smaller and smaller views; you already have that the different views are consistent on the overlaps; there are details that you can only see once you are "sufficiently close", and the inverse limit is a way of keeping track of all the details you can see at different resolutions. –  Arturo Magidin May 11 '11 at 19:47
I just have to remark that I have to give a general lecture in some seminar about forcing, I was thinking about the description in "other mathematics" and had very similar description such as Cauchy sequences describing a real number, and so on... :-) –  Asaf Karagila May 11 '11 at 20:52
And will the downvoter say why the downvote? –  Arturo Magidin May 13 '11 at 16:57

Well, I can give you my understanding of inverse limits from a topological view. Topologically, I think of inverse limits as "generalized intersections".

Consider a sequence of inclusions

$X_1 \supset X_2 \supset X_3 \supset \dots$

with an inverse limit $X$. Then $X$ is precisely the intersection of the $X_i$. The key thing here is that a point in the intersection gives you a point $x_i$ in each of the $X_i$, such that $x_{i+1}$ gets mapped to $x_i$. This is kind of silly and clear since you're really talking about the same point, just in the context of different containing sets, but it sets the stage for the more general inverse limit.

More generally, you can have maps between the $X_i$ that are not inclusions.

$X_1 \leftarrow X_2 \leftarrow X_3 \leftarrow \dots$

In this case, you can build an intersection-like object by taking a "point" in it to be a set of points, one from each $X_i$, such that each one is mapped to the previous. A better way to describe this is as a subset of the product $\Pi X_i$, namely points $(x_1, x_2, ...)$ such that $x_{i+1}$ maps to $x_i$. This also gives us a handy topology for this set, namely the induced topology from the product topology.

So say, for example, that $X_1$ is a point, $X_2$ is two (discrete) points, $X_3$ is four discrete points, etc., and each map is some two-to-one map. Then since each point has exactly two preimages, a point in the inverse limit is basically a choice of preimage (say, 0 or 1) at each $i$. The product topology makes these choices "close" if they agree for a long time, and so you may be able to then understand why the inverse limit is the Cantor set.

Basically, I think of the inverse limits as "infinite paths of preimages", with a topology that makes two sequences close if they stay close to each other for a long time.

A fun exercise: Let all the $X_i$ be circles, and let the maps from $X_{i+1}$ to $X_i$ be degree 2 maps (say, squaring the complex numbers with modulus 1). Understand the inverse limit of this system (it's often called a "solenoid").

Hopefully that helps you get a different perspective!

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I am grateful for a topological example, in a crazy land of categories and algebra. Feels very homey :-) However two things are unclear here, what does "a topology that makes two sequences close if they stay close to each other for a long time" mean in actual terms? in the fun exercise are we taking any circles (and what is a degree 2 map anyway?) –  Asaf Karagila May 12 '11 at 22:51
Sorry about the hand-waveyness. The inverse limit $X$ is a subset of the product of the $X_i$, and its topology is simply the subspace topology induced by the product topology. So when is a sequence $a_1, a_2, a_3 , ... $ in $X$ convergent? Well remember, each $a_k$ is itself a sequence, with a chosen point in each $X_i$. The $a_k$ converge if, for each fixed $i$, the chosen point in $X_i$ is the same for all large enough $k$. To make the fun exercise more concrete: Let each $X_i$ be a copy of the unit circle in $\mathbb{C}$, and let each map be $z\rightarrow z^2$. –  MartianInvader May 17 '11 at 21:03
To even better clarify "paths of preimages", each point in the inverse limit is basically choosing a point in some $X_i$, choosing a preimage of that point in $X_{i+1}$, a preimage of that point in $X_{i+2}$, etc. Two such sequences are "close" if they have the same choices up to some big $i$. Also, I guess I should mention that inverse limits can be pretty horrible in topology. Even something as simple as the squaring example I gave gives an inverse limit with uncountably many path components. –  MartianInvader May 17 '11 at 21:32
@MartianInvader, "The inverse limit X is a subset of the product of the X_i". Since a (2-valued) binary relation in set theory is a subset of products of two sets, then is an inverse limit an infinite-arity relation? –  alancalvitti Dec 6 '14 at 15:16

Here's how I think of inverse versus direct limits. The following should not be taken too seriously, because of course inverse limits and colimits are completely dual. But in many of the categories one tends to work with daily, they have a somewhat different feel.

In practice, inverse limits are a lot like completions: one has a tower of spaces $X_n \to X_{n-1} \to X_{n-2} \to \dots$ and one wishes to consider the space of "Cauchy sequences": in other words, one has a sequence $(x_n)$ such that $x_n \in X_n$ (this is the Cauchy sequence) and the $x_n$ are compatible under the maps (which is the abstract form of Cauchy-ness). For instance, the completion of a ring is the standard example of an inverse limit in commutative algebra. Here the $X_n$'s (which are the quotients of a fixed ring $R$ by a descending sequence of ideals $I_n$) can be thought of as specifying sets of "intervals" that are shrinking with each $n$, so an element of the inverse limit is a descending sequence of intervals. Perhaps the reason that inverse limits feel this way in many categories of interest is that many categories of interest are concrete, and the forgetful functor to sets is corepresentable, so that (categorical) limits look the same in the category and in the category of sets. In this case, inverse limits are given by precisely the construction above: it is a kind of "successive approximation."

Direct limits, on the contrast, are much more like unions. Here the picture that I usually keep in mind is that of a sequence of objects in time that gets wider as time passes, even though this is not necessarily accurate: in practice, one often wishes to take direct limits over non-monomorphisms. But the construction of a direct limit in the category of sets (and in many concrete categories: often, it happens that the forgetful functor also commutes with direct limits, and one deeper reason for that is that the corepresenting object is relatively small, and small objects have a tendency to commute with filtered colimits because of the above union interpretation) is ultimately the quotient of the disjoint union of the sets such that each element of a set $X_n$ is identified with its image in the next one. Since ultimately it feels like taking a union, direct limits tend to behave very nicely homologically: most often, they preserve exactness. Inverse limits, by contrast, do not usually preserve exactness unless one imposes extra conditions (such as the ML condition).

Finally, as a categorical limit, inverse limits are always easy to map into, while direct limits are always easy to map out of.

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I too think of the Cauchy sequences analogy. This was suggested to me by D.N. Yetter. Good answer +1. Especially the last line, it is catchy and mnemonous. –  BBischof May 11 '11 at 18:36
I am very glad to see you taking part of the website again. I understand some of your answer, but I want to give it some more thought before asking specific questions. –  Asaf Karagila May 11 '11 at 20:12
After reading Arturo's answer in details your answer still seems to elude from my grasp into the realm of algebraic and categorical definitions. I am aware that I am asking to clean out the categoricity from a very categorical construction, I have lost you after the completion example, and again after two lines in the second paragraph. –  Asaf Karagila May 12 '11 at 22:46
@Asaf: Dear Asaf, thank you for the kind words, and I'm sorry my answer wasn't more helpful. I'm not quite sure that I know much more to say (especially which the other answers haven't covered quite nicely!). Cheers, –  Akhil Mathew May 13 '11 at 14:36
Do not worry, judging by the +15 (at time of commenting) score of your answer, even if it is the lowest score amongst these answers it still signals that it is a very good answer. I am only grateful my question had the luxury of three wonderful answers that will certain help other people someday! –  Asaf Karagila May 13 '11 at 17:23

This has puzzled me for a long time, until I came across the following example that should appeal to a set theorist:

Let $S$ be a set and let $(\Pi_n)$ be a sequence of partitions of $S$ such that $\Pi_{n+1}$ is finer than $\Pi_n$ for all $n$. That is, every cell in $\Pi_n$ is the union of cells in $\Pi_{n+1}$. Now for each $n\geq m$, $\Pi_n$ is finer than $\Pi_m$ and there is a canonical projection $\pi_{mn}:\Pi_n\to\Pi_m$ that maps each cell in $\Pi_n$ to the unique cell in $\Pi_m$ that contains the former as a subset.


The underlying directed set is simply $\mathbb{N}$ with the natural order. The partitions together with the projections form a projective system:

  1. $\pi_{nn}$ is the identity on $\Pi_n$.
  2. $\pi_{ln}=\pi_{lm}\circ\pi_{mn}$ if $l\leq m\leq n$.

So what is the projective limit? By definition, it is $$\Big\{(P_0,P_1,P_2,\ldots)\in\prod_{n=0}^\infty\Pi_n:P_m=\pi_{mn}(P_n)\text{ for }m\leq n\Big\}.$$

Since all (non-identity-)projections can be made up from projections of the form $\pi_{n(n+1)}$, we can rewrite this as


$$=\Big\{(P_0,P_1,P_2,\ldots)\in\prod_{n=0}^\infty\Pi_n:P_n\supseteq P_{n+1}\Big\}.$$

So the inverse limit is simply the set of infinite paths in the tree formed from this sequence of partitions.

There a related approaches in probability theory and measure theory as approaches to generalize the Daniell-Kolmogorov extension Theorem. Studying projective limits of probability spaces was pioneered by Salomon Bochner. A typical paper in this strand of literature is this paper by Kazimierz Musiał.

A beautiful (and very short) paper that gives an example in which the projective limit is empty even though all projections are surjective is this note by William Waterhouse.

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This is a very clear example for me! Just to be clear: $P_i$ is a "cell" in partition $\Pi_i$, correct? –  magma May 14 '12 at 21:50
Yes, $P_i$ is a "cell" or "block" in the partition. –  Michael Greinecker May 14 '12 at 21:51
thank you Michael. –  magma May 14 '12 at 22:26

I'm currently dabbling around in a category where morphisms are refinements of computer programs, in the sense that they are functions $f : X \rightarrow Y$ from a program refinement (an implementation) $X$ to a program specification $Y$.

In this setting, a sequence (or net) of morphisms can be either describing an ever more elaborate specification $X_0 \rightarrow X_1 \rightarrow X_2 \rightarrow ...$ or it can be an ever more precise refinement $... \rightarrow X_2 \rightarrow X_1 \rightarrow X_0$. The first sequence has as a direct limit the 'ultimate' specification, while the second sequence has as an inverse (also called projective) limit the 'ultimate' refinement.

As an extra: maybe my own question about Directed and projective limit in Rel helps in looking at the notion in a different way. In Rel, where the two types of limit coincide, my intuition is that a net of relations simply results in 'lines' connecting points in a disjoint union, and the 'infinite' lines form a new object, the limit.

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I like the picture of a cocone from http://arxiv.org/abs/math/0306223:


They describe it through a story about people sending emails to each other.

And then:

A cocone which is universal is a colimit. ―http://ncatlab.org/nlab/show/cocone


Colimits glue. Limits cut. ―Adam Hughes

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I hate diagrams, they never help and they almost always confuse. But I suppose other people might find this helpful. So thanks! –  Asaf Karagila May 22 at 6:16
Indeed, this is how I was presented colimits for the first time. Paraphrasing one of my professors, the appropriate (and possibly only) way to work with this definition is to "throw universal properties at it". This might not give you much intuition into what you're actually doing (at least, it didn't give me any). It took me a long while to understand ring completions... –  A.P. May 22 at 9:23
@AsafKaragila the claim that diagrams never help is uncharacteristically absurd coming from you... –  Mariano Suárez-Alvarez May 23 at 2:58
@Mariano: How so? –  Asaf Karagila May 23 at 5:10
@Mariano: I hope that it's clear that I am only talking for myself here; as was somewhat evident by the second half of my comment that other people might find it useful. And I can't recall such time where diagram simplified something for me, at best they didn't clarify the written definition, but more often than not, they end up confusing me compared to a well-written definition. –  Asaf Karagila May 23 at 13:12

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