# The Unapologetic Mathematician

## Limits of sets and Creation of limits

We know that the category $\mathbf{Set}$ has equalizers and all (small) products, so by the existence theorem we know it is complete. However, it will be useful to have an explicit calculation of all small limits at hand. We’ll find this by walking through the existence theorem in this category.

Given any small category $\mathcal{J}$, a functor $F:\mathcal{J}\rightarrow\mathbf{Set}$ is a collection of sets $F(J)$ and functions $F(u):F(J)\rightarrow F(K)$. We can take the product of all the sets in the collection $\prod_JF(J)$, consisting of all lists of elements $x=\{x_J\}_{J\in\mathrm{Ob}(\mathcal{J})}$ with each $x_J\in F(J)$. It comes with projections $\pi_J:\prod_KF(K)\rightarrow F(J)$, but the triangles formed by these projections might not commute — $\pi_K$ is not equal to $F(u)\circ \pi_J$ in general.

So take the subset $L$ of the product consisting of all those lists $\{x_J\}_{J\in\mathrm{Ob}(\mathcal{J})}$ with $\left[F(u)\right](x_J)=x_K$ for all morphisms $u$ of $\mathcal{J}$. Then we have functions $\lambda_J:L\rightarrow F(J)$ defined by $\lambda_J(x)=x_J$ which do make the required triangles commute, giving a cone on $F$. If $\mu_J:M\rightarrow F(J)$ is any other cone on $F$ then each $m\in M$ determines a list $x(m)=\{\mu_J(m)\}$ which lies in $L$, which determines a unique function $x:M\rightarrow L$.

Now such a list $\{x_J\}$ is a list of elements of the various $F(J)$, and when it comes right down to it we just don’t like thinking about elements if we don’t have to. Instead of talking about an element $x$ of a set $X$, we can talk about an arrow $x:*\rightarrow X$, where ${*}$ is a terminal object in $\mathbf{Set}$ — a set with one point. So what we really have here is a list of arrows $x_J:*\rightarrow F(J)$ with $F(u)\circ x_J=x_K$, which is a cone on $F$!

So here’s the result of our calculation: The limit set $\varprojlim_\mathcal{J}F$ is the set $\mathrm{Cone}(*,F)$ of all cones from ${*}$ to $F$ in $\mathbf{Set}$. Further, we see that this set always exists, so $\mathbf{Set}$ is complete.

We could do a similar calculation in another category like $\mathbf{Grp}$, but we actually don’t have to reproduce all this work. Instead, we can consider the functor $U:\mathbf{Grp}\rightarrow\mathbf{Set}$ which assigns to every group its underlying set. If $F:\mathcal{J}\rightarrow\mathbf{Grp}$ is a functor, then the composite $U\circ F:\mathcal{J}\rightarrow\mathbf{Set}$ has a limit $L=\mathrm{Cone}(*,U\circ F)$. I claim that there is a unique group structure on this set so that the arrows $\lambda_J:L\rightarrow U(F(J))$ are actually group homomorphisms to the groups $F(J)$, and that this cone in $\mathbf{Grp}$ is a limiting cone.

Since the projection $\lambda_J:L\rightarrow F(J)$ just reads off the $J$ component from the list, we see that for it to be a homomorphism we have to use the product from $F(J)$. That is, $\lambda_J(\{x_J\}\{y_J\})=\lambda_J(\{x_J\})\lambda_J(\{y_J\})=x_Jy_J$, so the $J$ component of the product must be $x_Jy_J$, using the product from $F(J)$. Similarly the $J$ component of $\{x_J\}^{-1}$ must be $x_J^{-1}$, using the inversion from $F(J)$.

Now if $\gamma_J:G\rightarrow F(J)$ is any cone in $\mathbf{Grp}$ then $U(\gamma_J):U(G)\rightarrow U(F(J))$ is a cone in $\mathbf{Set}$, so $U(\gamma_J)=\lambda_J\circ h$ for some unique function $G\rightarrow L$. We can check that $\lambda_J(h(g_1g_2))=\gamma_J(g_1g_2)=\gamma_J(g_1)\gamma_J(g_2)=\lambda_J(h(g_1))\lambda_J(h(g_2))=\lambda_J(h(g_1)h(g_2))$

so $h$ must also be a group homomorphism, proving the required universal property.

In general we say that a functor $G:\mathcal{C}\rightarrow\mathcal{D}$ creates a limit for a functor $F:\mathcal{J}\rightarrow\mathcal{C}$ if for every limiting cone $\lambda_J:L\rightarrow G(F(J))$ in $\mathcal{D}$ there exists a unique cone $\mu_J:M\rightarrow F(J)$ in $\mathcal{C}$ with $L=G(M)$ and $\lambda_J=G(\mu_J)$, and further this cone is a limiting cone. The above theorem can then be stated that the functor $U:\mathbf{Grp}\rightarrow\mathbf{Set}$ creates all limits. As a consequence, $\mathbf{Grp}$ is complete.

A similar proof to the preceding theorem can be given to show that the “underlying set” functors from $\mathbf{Ab}$, $\mathbf{Ring}$, $R\mathbf{-mod}$, $\mathbf{mod-}R$, and other algebraic categories all create all (small) limits, and thus those categories are complete. This also shows that the group (ring, module, etc.) structure on a product (equalizer, pullback) is completely determined by the limit on the underlying sets.

June 21, 2007 - Posted by | Category theory

1. Absolutely fantastic work. I just got hooked to your writings, recently via nCatCafe.
I was wondering how you produce such fine diagrams and symbols . May be you explain in one of your older posts..If not, I would much like to know your blog entry creation process.
-cheers
raj Comment by Raj | June 22, 2007 | Reply

2. The inline LaTeX is explained here. It also works for comments, and it’s good for all WordPress blogs, so if you want to use it in comments at Borcherds’ or Tao’s place, or at the Secret Blogging Seminar, that should do it.

For the larger diagrams I use Paul Taylor’s diagrams package in my actual TeX editor, then I capture that area of the screen, crop it to a nice round size, convert to JPEG, and upload as an image. Comment by John Armstrong | June 22, 2007 | Reply

3. […] of limits Yesterday I talked about functors that create limits. It’s theoretically interesting, but a different condition that’s often more useful in […]

Pingback by Preservation of limits « The Unapologetic Mathematician | June 22, 2007 | Reply

4. […] in functor categories Today I want to give a great example of creation of limits that shows how useful it can be. For motivation, take a set , a monoid , and consider the set of […]

Pingback by Limits in functor categories « The Unapologetic Mathematician | June 23, 2007 | Reply

5. 2 (minor) typoes and a query.

par 5: from to F => from * to F

par 8: some unique function G->L => some unique function h:G->L

Query: Is the uniqueness part of the definition of “creates a limit” needed for getting the proof that GRP is complete? I can’t see why. Comment by Avery | February 12, 2008 | Reply

6. On the subject of LaTeX errors, whenever you write * as a single LaTeX image, it seems to not show up (at least not on the old version of IE I’m using at the mo, and I think not on firefox either). So eg.
So here’s the result of our calculation: The limit set (LaTeX image) is the set (LaTeX image) of all cones from (no visible LaTeX image) to… Comment by PhiJ | March 6, 2008 | Reply

7. Thanks PhiJ. That’s just the sort of thing that worked when I created it but got broken later. Comment by John Armstrong | March 6, 2008 | Reply

8. […] space. This is a forgetful functor, and the usual abstract nonsense can be used to show that it creates limits. And from there it’s straightforward to check that the category of inner product spaces is […]

Pingback by The Category of Inner Product Spaces « The Unapologetic Mathematician | May 6, 2009 | Reply