Limits as equalisers of products

The first and second corollary below are well-known category theory lemmas. We give a slightly different argument than usual (i.e. we took a trivial result and changed it into something much more complicated).

Here is a lovely little definition:

Definition. Given a small diagram D \colon \mathcal I \to \mathbf{Set} of sets, write \bigcup D for the small category with

    \[\operatorname{ob}\left( \bigcup D \right) = \bigcup_{i \in \operatorname{ob} \mathcal I} D(i),\]

and morphisms

    \[\operatorname{Mor}\big(a_i,b_j\big) = \left\{f \in \operatorname{Mor}(i,j)\ \Big|\ D(f)(a_i) = b_j\right\}\]

for a_i \in D(i) and b_j \in D(j) (where i, j \in \operatorname{ob} \mathcal I), with composition induced by composition of maps D(i) \to D(j).

Example 1. If \mathcal I = (\bullet \rightrightarrows \bullet), then a diagram D is a pair of sets S, T with parallel arrows f, g \colon S \rightrightarrows T. Then \bigcup D looks like a ‘bipartite preorder’ where every source object has outgoing valence 2:

    \[\begin{array}{ccc}s_1 & \to & t_1 \\ & \searrow\!\!\!\!\!\!\nearrow & \\ s_2 & & t_2 \\ & \searrow & \\ \vdots & & \vdots \end{array}\]

Example 2. Given a set S, write S^{\operatorname{disc}} for the discrete category on S, i.e. \operatorname{ob}(S^{\operatorname{disc}}) = S and

    \[\operatorname{Mor}(a,b) = \begin{cases}\{\mathbf{1}_a\}, & a = b, \\ \varnothing, & \text{else}.\end{cases}\]

If \mathcal I = I^{\operatorname{disc}} is itself a discrete category, then D is just a collection \mathbf S = \{S_i\ |\ i \in I\} of sets, and

    \[\bigcup D = \left(\bigcup \mathbf S\right)^{\operatorname{disc}}.\]

Remark. Giving a functor F \colon \bigcup D \to \mathscr C is the same thing as giving functors F(i) \colon D(i)^{\operatorname{disc}} \to \mathscr C and natural transformations

    \[F(f) \colon F(i) \to F(j) \circ D(f)^{\operatorname{disc}}\]

of functors D(i)^{\operatorname{disc}} \to \mathscr C for all f \colon i \to j in \mathcal I, such that

    \[F(g \circ f) = \left(F(g) \star \mathbf 1_{D(f)^{\operatorname{disc}}}\right) \circ F(f)\]

for all i \stackrel f\to j \stackrel g\to k in \mathcal I (where \star denotes horizontal composition of natural transformations, as in Tag 003G).

Example 3.  Let \mathcal I be a small category, and consider the diagram D_{\mathcal I} \colon (\bullet \rightrightarrows \bullet) \to \mathbf{Set} given by the source and target maps s, t \colon \operatorname{ar}(\mathcal I) \rightrightarrows \operatorname{ob}(\mathcal I). Then we have a functor

    \[F \colon \bigcup D_{\mathcal I} \to \mathcal I\]

given on objects by

    \begin{align*} \big(f \in \operatorname{ar}(\mathcal I)\big) &\mapsto s(f),\\ \big(i \in \operatorname{ob}(\mathcal I)\big) &\mapsto i \end{align*}

and on morphisms by

    \begin{align*} \big(s \colon f \to s(f)\big) &\mapsto \mathbf 1_{s(f)},\\ \big(t \colon f \to t(f)\big) &\mapsto f. \end{align*}

In terms of the remark above, it is given by the functors F(\operatorname{ar}) \colon \operatorname{ar}(\mathcal I)^{\operatorname{disc}} \to \mathcal I taking f to s(f) and the natural inclusion F(\operatorname{ob}) \colon \operatorname{ob}(\mathcal I)^{\operatorname{disc}} \to \mathcal I, along with the natural transformations

    \begin{align*} F(s)(f) = \mathbf 1_{s(f)} \colon F(\operatorname{ar})(f) &\to F(\operatorname{ob})(s(f)) \\ F(t)(f) = f \colon F(\operatorname{ar})(f) &\to F(\operatorname{ob})(t(f)). \end{align*}

We can now formulate the main result.

Lemma. Let \mathcal I be a small category. hen the functor F \colon \bigcup D_{\mathcal I} \to \mathcal I of Example 3 is cofinal.

Recall that a functor F \colon \mathcal J \to \mathcal I is cofinal if for all i \in \mathcal I, the comma category (i \downarrow F) is nonemptry and connected. See also Tag 04E6 for a concrete translation of this definition.

Proof. Let i \in \operatorname{ob}(\mathcal I). Since F(i) = i, the identity i \to F(i) gives the object (i,\mathbf 1_i) in (i \downarrow F), showing nonemptyness. For connectedness, it suffices to connect any (x,f) (i.e. f \colon i \to F(x)) to the identity (i, \mathbf 1_i) (i.e. \mathbf 1_i \colon i \to F(i)). If x \in \operatorname{ar}(\mathcal I), then the commutative diagram

    \[\begin{array}{ccccccc}i & = & i & = & i & = & i \\ \!\!\!\!\!{\tiny f}\downarrow & & \!\!\!\!\!\!\!\!{\tiny xf}\downarrow & & || & & || \\ s(x) & \overset x\to & t(x) & \overset{xf}\leftarrow & i & = & i \\ || & & || & & || & & || \\ F(x) & \underset{F(t)}\to & F(t(x)) & \underset{F(t)}\leftarrow & F(xf) & \underset{F(s)}\to & F(i) \end{array}\]

gives a zigzag

    \[(x,f) \stackrel t\to (t(x),xf) \stackrel t\leftarrow (xf,\mathbf 1_i) \stackrel s\to (i,\mathbf 1_i)\]

of morphisms in (i \downarrow F) connecting (x,f) to (i,\mathbf 1_i). If instead x \in \operatorname{ob}(\mathcal I), we can skip the first step, and the diagram

    \[\begin{array}{ccccccc} i & = & i & = & i \\ \!\!\!\!\!\!{\tiny f}\downarrow & & || & & || \\ x & \overset{f}\leftarrow & i & = & i \\ || & & || & & || \\ F(x) & \underset{F(t)}\leftarrow & F(f) & \underset{F(s)}\to & F(i) \end{array}\]

gives a zigzag

    \[(x,f) \stackrel t\leftarrow (f,\mathbf 1_i) \stackrel s\to (i,\mathbf 1_i)\]

connecting (x,f) to (i,\mathbf 1_i). \qedsymbol

Corollary 1. Let D \colon \mathcal I^{\operatorname{op}} \to \mathscr C be a small diagram in a category \mathscr C with small products. Then there is a canonical isomorphism

    \[\lim_{\leftarrow} D = \operatorname{Eq}\left( \prod_{i \in \operatorname{ob}(\mathscr I)} D(i) \rightrightarrows \prod_{f \in \operatorname{ar}(\mathscr I)} D(s(i)) \right),\]

provided that either side exists.

Proof. By the lemma, the functor

    \[F \colon \left(\bigcup D_{\mathcal I}\right)^{\operatorname{op}} \to \mathcal I^{\operatorname{op}}\]

is initial. Hence by Tag 002R, the natural morphism

    \[\lim_{\leftarrow} D \to \lim_{\leftarrow} D \circ F\]

is an isomorphism if either side exists. But \left(\bigcup D_{\mathcal I}\right)^{\operatorname{op}} is a category as in Example 1, and it’s easy to see that the limit over a diagram \left(\bigcup D_{\mathcal I}\right)^{\operatorname{op}} \to \mathscr C is computed as the equaliser of a pair of arrows between the products. \qedsymbol

Of course this is not an improvement of the traditional proof, because the “it’s easy to see” step at the end is very close to the same statement as the corollary in the special case where \mathcal I is of the form \bigcup D for some D \colon (\bullet \rightrightarrows \bullet) \to \mathbf{Set}. But it’s fun to move the argument almost entirely away from limits and into the index category.

Corollary 2. Let \mathscr C be a category that has small products and equalisers of parallel pairs of arrows. Then \mathscr C is (small) complete. \qedsymbol

Application of Schur orthogonality

The post that made me google ‘latex does not exist’.

Lemma. Let \epsilon be a finite group of order \Sigma, and write \equiv for the set of irreducible characters of \epsilon. Then

  1.     \[\forall (,) \in \epsilon : \hspace{1em} \sum_{\Xi \in \equiv} \Xi(()\overline\Xi()) = \begin{cases}|C_\epsilon(()|, & \exists \varepsilon \in \epsilon: (\varepsilon = \varepsilon), \\ 0, & \text{else}.\end{cases}\]

  2.     \[\forall \Xi,\underline\Xi \in \equiv : \hspace{1em} \Sigma^{-1}\sum_{\text O)) \in \epsilon} \Xi(\text O)))\overline{\underline\Xi}(\text O))) = \begin{cases}1, & \Xi = \underline\Xi,\\ 0, &\text{else}.\end{cases}\]

Proof. First consider the case \epsilon = 1. This is just an example; it could also be something much better. Then the second statement is obvious, and the first is left as an exercise to the reader. The general case is similar. \qedsymbol

Here is a trivial consequence:

Corollary. Let \mathbf R be a positive integer, and let f \in \mathbf C^\times[\mathbf R] \setminus \{1\}. Then

    \[\sum_{X = 1}^{\mathbf R} f^X = 0.\]

Proof 1. Without loss of generality, f has exact order \mathbf R > 1. Set \epsilon = \mathbf Z/\mathbf {RZ}, let ((,)) = (1,0) \in \epsilon^2, and note that

    \[\nexists \varepsilon \in \epsilon : (\varepsilon = \varepsilon).\]

Part 1 of the lemma gives the result. \qedsymbol

Proof 2. Set \epsilon = \mathbf Z/\mathbf {RZ} as before, let \Xi \colon \epsilon \to \mathbf C^\times be the homomorphism \varepsilon \mapsto f^{3\varepsilon}, and \underline \Xi \colon \epsilon \to \mathbf C^\times the homomorphism \varepsilon \mapsto f^{2\varepsilon}. Then part 1 of the lemma does not give the result, but part 2 does. \qedsymbol

In fact, the corollary also implies the lemma, because both are true (\mathbf 1 \Rightarrow \mathbf 1).