# Colimits glue. What do limits do?

The informal but very useful way to think of colimits is as 'gluing things'. This intuitive view is very helpful to me, and I'd like one for limits aswell, but I haven't stumbled opon such a thing anywhere so...

If colimits glue, what do limits do?

• They "cut" things. The idea of a limit from sets is intersection, so you should think of it as cutting off bits in that sense. It's a subset of a product, so it makes sense that you're cutting off the ways they are incompatible (transition functions). The main problem is that colimits take coproducts, which are like assembling pieces to glue, but limits start with products, which somewhat intertwines how they work together, so the metaphor is less clear in this case. – Adam Hughes Jan 7 '15 at 23:13
• This intuition is all well and good but it's very easy to find cases where it doesn't apply. – Matt Samuel Jan 7 '15 at 23:19
• I think it's probably better to link en.wikipedia.org/wiki/Inverse_limit (limit) and en.wikipedia.org/wiki/Direct_limit (colimit). There are examples there. – Matt Samuel Jan 7 '15 at 23:32
• They coglue, obviously. – Najib Idrissi Jan 8 '15 at 13:19
• @NajibIdrissi was waiting for that :p – user153312 Jan 8 '15 at 15:07

Limits cut out solutions to equations.

Edit: For example, this is literally true in the case of affine schemes. Consider the affine scheme $\mathbb{A}^n = \text{Spec } k[x_1, \dots x_n]$ over a field $k$, and let $f(x_1, \dots x_n) \in k[x_1, \dots x_n]$ be a polynomial. $f$ defines a map $\mathbb{A}^n \to \mathbb{A}^1$, and the scheme $V(f) = \text{Spec } k[x_1, \dots x_n]/f$ of solutions to $f$ is precisely the fiber of $f$ over $0 \in \mathbb{A}^1$, which is a special case of the pullback.

• Possibly infinitely many equations in possibly infinitely many variables, which is kind of exotic. – tcamps Jan 8 '15 at 21:46
• Perhaps it would be helpful to give an example? – PyRulez Jan 8 '15 at 23:23
• An example, and an elaboration, would be nice. I don't understand what is meant by the answer out all. – user153312 Jan 8 '15 at 23:51
• @Exterior: sorry, I was in a hurry when I first wrote it. Edited with an example. – Qiaochu Yuan Jan 9 '15 at 6:22

As an example of limits cutting out solutions to equations think of equalizers or pullbacks in $\mathsf{Set}$. Do it now, before reading on. This is the sort of example that you need to have at your fingertips.

In any complete category $\mathcal{C}$, the limit of a functor $F: \mathcal{I} \to \mathcal{C}$ can be computed as the equalizer of two maps $\prod_{\alpha \in \mathbf{Mor} \mathcal{I}}F(\mathrm{dom}\alpha) \overset{\to}{\underset{\to}{}} \prod_{i \in I} Fi$. In $\mathsf{Set}$, the two arrows are functions in $|\mathbf{Mor}\mathcal{I}|$ variables, and the equalizer is the set of all solutions to setting these two functions equal to each other.

Now remember that the categories you're thinking of are concrete categories, usually admitting a limit-preserving (even right adjoint) functor to $\mathsf{Set}$. So their limits are calculated in the same way, with some bells and whistles to specify the structure forgotten by passing to the underlying set.

Of course, these kinds of intuitions have their limitations -- most obviously, by passing to the opposite of the category you're considering, you switch limits and colimits.

Even in perfectly ordinary settings, sometimes the "cutting-out-by-equations" notion is not the easiest way (for me) to think about limits. For instance, consider the $p$-adic integers $\mathbb{Z}_p = \varprojlim \mathbb{Z}/p^n$ where the limit is taken in $\mathsf{Ab}$ over the obvious chain of quotient maps $\dots \to \mathbb{Z}/p^{n+1} \to \mathbb{Z}/p^n \to \dots$. The cut-out-by-equations description describes this as the subgroup of $\alpha \in \prod_n \mathbb Z/p^n$ such that for each $n$, $\alpha_n = p\alpha_{n+1}$. I tend to think of this as the coordinates $\alpha_n$ being "glued" together -- somehow the individual elements of the limit are glued together whereas in a colimit it's the whole space itself that's been glued together. This can even be seen in the case of pullbacks in $\mathsf{Set}$: an element of the pullback can be thought of as "glued together" from elements of the two upstairs sets, "glued together" by an equation between them downstairs. I'm not quite sure if this can be explained in some categorical way. Maybe (homotopy?) type theory could help to understand this.

I also tend not to think of $\mathbb Z_p$ as being "cut out" of $\prod_n \mathbb{Z}/p^n$ even though it is: I think the reason is that the latter space is so huge and $\mathbb{Z}_p$ is so small -- it is a subspace of very high codimension. I suppose this is no different from the fact that I don't usually explicitly think of a CW complex as a quotient of the coproduct of all its cells.

• I'm curious about you comment on CW complexes. Isn't the all point of CW complexes to be able to think of certain topological spaces as the glueing of cells to have a combinatorial intuition about them ? – Pece Jan 9 '15 at 6:35
• Yeah, I guess all I'm saying is that I could visualize the big coproduct space of which my complex is a quotient as a sort of "étale cover" of my complex, but I usually don't. Similarly, when thinking about $\mathbb Z_p$, I rarely visualize the big product space $\prod_n \mathbb Z / p^n$ with $\mathbb Z_p$ sitting inside. – tcamps Jan 9 '15 at 13:13
• Some of the answers to this question get at the idea that limits can be "completions", which don't feel like generic "solution sets of equations" to me. But maybe the way to look at it is that it's an interesting insight that "completions" really are "solution sets of equations"... – tcamps Feb 11 '15 at 0:09

Sheaves provide a nice framework for getting comfortable with the notion of a limit. See my answer at the relevant MO discussion

MO/23268: Geometric intuition for limits

There you will also find many other interesting answers.

To add something "new" and explain a little bit the other answers: Compute the pullback of the two obvious maps $(S^2)^+ \to S^2 \leftarrow (S^2)^-$, where by $(S^2)^+$ I mean the upper hemisphere and by $(S^2)^-$ the lower hemisphere.