Ends Coends and Weighted Limits — Examples

The integrand is $H(j_1,j_2)=\mathbf{S}\mathbf{e}\mathbf{t}(F(j_1),G(j_2))$, with values $H(0,0)$ of size four, $H(1,1)$ of size two, $H(0,1)$ of size four and $H(1,0)$ of size two. The object-indexed product is $\mathbf{S}\mathbf{e}\mathbf{t}(F(0),G(0))\times \mathbf{S}\mathbf{e}\mathbf{t}(F(1),G(1))$, with eight elements; the morphism-indexed product has one factor for each of $1_0$, $1_1$ and $u$, namely $\mathbf{S}\mathbf{e}\mathbf{t}(F(0),G(0))$, $\mathbf{S}\mathbf{e}\mathbf{t}(F(1),G(1))$ and $\mathbf{S}\mathbf{e}\mathbf{t}(F(0),G(1))$.

By [L2] the two parallel maps send a pair $({\varphi }_0,{\varphi }_1)$ to the families whose $f$-component is $G(f)\circ {\varphi }_{\mathrm{dom}f}$ and ${\varphi }_{\mathrm{cod}f}\circ F(f)$ respectively. At $1_0$ and at $1_1$ both components are ${\varphi }_0$ and ${\varphi }_1$, so those factors impose nothing; at $u$ the condition is $G(u)\circ {\varphi }_0={\varphi }_1\circ F(u)$. Since $G(u)$ sends both $a$ and $b$ to $c$, the left side is the constant function at $c$, and the right side is the constant function at ${\varphi }_1(r)$; so the condition holds exactly when ${\varphi }_1(r)=c$. The equalizer is therefore the subset of the eight-element product on which ${\varphi }_1(r)=c$, which leaves ${\varphi }_0$ free: its elements are the four pairs with ${\varphi }_0$ one of $p,q\mapsto a,a$; $a,b$; $b,a$; $b,b$ and ${\varphi }_1(r)=c$.

Listing the natural transformations independently gives the same four. By [F2] a natural transformation $\alpha :F\Rightarrow G$ is a pair ${\alpha }_0:\{p,q\}\to \{a,b\}$ and ${\alpha }_1:\{r\}\to \{c,d\}$ with $G(u){\alpha }_0={\alpha }_{1}F(u)$, which as in step 2.1 says exactly ${\alpha }_1(r)=c$; there are four choices of ${\alpha }_0$ and each extends in exactly one way.

The two lists agree pair for pair, and by [L1] they must: the end of $H$ is $\mathrm{Nat}(F,G)$ with the evaluation wedge, and by [L2] it is the equalizer computed in step 2.1. So the end has four elements on this diagram, and the identification is the identity on the four pairs.

The assignment $S$ is a functor: for $f:B\to B^{\mathrm{\prime }}$ the contravariant slot acts by $g\mapsto g\circ f$ from $Y^{B^{\mathrm{\prime }}}$ to $Y^B$ and the covariant slot by $f$ itself, and both actions preserve identities and composites because composition of functions does. The family ${\mathrm{ev}}_B$ is the transpose $\check{k}$ of the identity of $Y^B$ under the bijection of [L1] with $A=B$ and $X=Y^B$, so it is the counit of that adjunction at $Y$.

Fix $f:B\to B^{\mathrm{\prime }}$ and chase an arbitrary element $(g,b)$ of $S(B^{\mathrm{\prime }},B)=Y^{B^{\mathrm{\prime }}}\times B$ through both legs of the hexagon. Since the target is the constant functor at $Y$, both outer actions on the target side are identities, and the two legs are ${\mathrm{ev}}_B\circ S(f,1_B)$ and ${\mathrm{ev}}_{B^{\mathrm{\prime }}}\circ S(1_{B^{\mathrm{\prime }}},f)$. The first sends $(g,b)$ to ${\mathrm{ev}}_B(g\circ f,b)=g(f(b))$ and the second sends it to ${\mathrm{ev}}_{B^{\mathrm{\prime }}}(g,f(b))=g(f(b))$. The two values agree for every $(g,b)$, so the two legs are equal.

Since the target is the constant functor at $Y$, the equation verified in step 2.1 is exactly the cowedge equation of [F2], so the evaluation family is a cowedge under $S$ with vertex $Y$, and in particular a dinatural transformation.

The objects of $\mathrm{Tw}(\mathcal{C})$ are the three morphisms $1_0$, $1_1$ and $u$ of $\mathcal{C}$, by [F1].

The morphisms are enumerated by testing, for each ordered pair of objects, whether the required factorisation exists. A morphism $1_0\to u$ needs $a:0\to 0$ and $b:0\to 1$ with $b\circ 1_0\circ a=u$, and the pair $(1_0,u)$ works and is the only one available. A morphism $1_1\to u$ needs $a:0\to 1$ and $b:1\to 1$ with $b\circ 1_1\circ a=u$, and the pair $(u,1_1)$ works and is the only one available. A morphism out of $u$ to either identity needs a component in $\mathcal{C}(1,0)$, which is empty, so there is none; a morphism $1_0\to 1_1$ needs a component in $\mathcal{C}(1,0)$ as well, and a morphism $1_1\to 1_0$ needs $b:1\to 0$. So apart from identities there are exactly the two morphisms named, both into $u$, and $\mathrm{Tw}(\mathcal{C})$ is a cospan.

The composite $T\pi$ takes the value $T(0,0)$ at $1_0$, the value $T(1,1)$ at $1_1$ and the value $T(0,1)$ at $u$, and it sends the morphism $(1_0,u)$ to $T(1_0,u)$ and the morphism $(u,1_1)$ to $T(u,1_1)$. By [L1] and [F3] the end of $T$ is the limit of that diagram, which by [F2] is exactly the pullback of the displayed cospan.

The category ${\mathcal{C}}_M$ has one object and its morphism collection is the set $M$, so it is small; by [F1] the integrand is $T(\ast ,\ast )=X\times Z$, and every value of $T$, on or off the diagonal, is that same set. The disjoint union over the objects therefore has one summand and is $X\times Z$ itself.

The generating pairs of [L1] are indexed by a morphism $m$ of ${\mathcal{C}}_M$, that is by an element of $M$, and by an element of the off-diagonal value, that is by a pair $(x,z)\in X\times Z$. The first leg $T(m,1_{\ast })$ acts by $P(m)$ in the contravariant slot and by the identity in the covariant one, giving $(x\cdot m,z)$; the second leg $T(1_{\ast },m)$ acts by the identity in the contravariant slot and by $F(m)$ in the covariant one, giving $(x,m\cdot z)$. So the generating pairs are exactly $(x\cdot m,z)$ against $(x,m\cdot z)$.

By [L1] and [F3] the coend is the quotient of the single summand $X\times Z$ by the least equivalence relation containing those pairs, which is the displayed description of $P{\otimes }_{{\mathcal{C}}_M}F$. At $m$ the identity of $M$ the two legs agree, so the identity contributes only reflexive pairs.

Take $T=\mathcal{C}(-,-)$, a functor into $\mathbf{S}\mathbf{e}\mathbf{t}$ by [L2], with $T(c,c)=\mathcal{C}(c,c)$ and off-diagonal value $T(c^{\mathrm{\prime }},c)=\mathcal{C}(c^{\mathrm{\prime }},c)$. By [F2] the two legs act on $x\in \mathcal{C}(c^{\mathrm{\prime }},c)$ as $T(f,1_c)(x)=1_c\circ x\circ f=x\circ f$ and $T(1_{c^{\mathrm{\prime }}},f)(x)=f\circ x\circ 1_{c^{\mathrm{\prime }}}=f\circ x$, using [F4]. So the generating pairs of [L1] are exactly the displayed ones, and the description of the coend follows from [L1] and [F3].

On the walking arrow, with objects $0$ and $1$ and one non-identity morphism $u:0\to 1$, the disjoint union is $\{(0,1_0),(1,1_1)\}$. A generating pair needs a morphism $f:c\to c^{\mathrm{\prime }}$ together with an element $x$ of $\mathcal{C}(c^{\mathrm{\prime }},c)$; the only non-identity morphism is $u:0\to 1$ and $\mathcal{C}(1,0)$ is empty, so it contributes none, while an identity $f=1_c$ gives $x\circ 1_c=1_c\circ x$ and hence only reflexive pairs. The relation is therefore equality and the coend has two elements.

On the one-object category ${\mathcal{C}}_M$ of a monoid $M$, the disjoint union has one summand and is $M$ itself, every value of the integrand being $M$; and by step 1.1 the generating pairs are $x\circ f$ against $f\circ x$ for $x,f\in M$. So the coend is $M$ modulo the least equivalence relation containing $xf\sim fx$. Nothing further is claimed for a general monoid.

If $M$ is a group, that quotient is the set of conjugacy classes. Each generating pair is a conjugation, since $fx=f(xf)f^{-1}$, and conjugacy is an equivalence relation, so the least equivalence relation containing the generators is contained in conjugacy; conversely, for $g$ and $h$ in $M$ the choice $x:=g{h}^{-1}$ and $f:=h$ gives $xf=g$ and $fx=hg{h}^{-1}$, so every conjugate pair is a generating pair and conjugacy is contained in the relation. The two therefore agree. Without inverses this second half is unavailable, which is why the general case of step 2.2 is stated only as the quotient by that relation.

Since $T$ ignores its two contravariant variables, [L2] applies with $\mathcal{K}=\mathcal{C}\times \mathcal{D}$ and $G(c,d)=F(c)\times F(d)$: the end over the product index category is the ordinary limit of $G$. A limit of a diagram on the walking arrow is the value at $0$, since a cone $({\lambda }_0,{\lambda }_1)$ over $A\to B$ satisfies ${\lambda }_1=h{\lambda }_0$ and is therefore determined by ${\lambda }_0$.

The limit of $G$ is $F(0)\times F(0)$, with four elements. Writing a cone with apex $X$ as ${\lambda }_{(c,d)}=({\mu }_{(c,d)},{\nu }_{(c,d)})$, the cone condition at $(1_c,u)$ gives ${\mu }_{(c,1)}={\mu }_{(c,0)}$ and ${\nu }_{(c,1)}=F(u){\nu }_{(c,0)}$, the condition at $(u,1_d)$ gives ${\nu }_{(1,d)}={\nu }_{(0,d)}$ and ${\mu }_{(1,d)}=F(u){\mu }_{(0,d)}$, and the condition at $(u,u)$ then holds automatically, both sides being $(F(u){\mu }_{(0,0)},F(u){\nu }_{(0,0)})$. So a cone is exactly a pair of functions $X\to F(0)$, and by [F2] the terminal one has apex $F(0)\times F(0)$.

The iterated ends give the same object. Holding the two $\mathcal{C}$-variables fixed at $(c_1,c_2)$ leaves the integrand $(d_1,d_2)\mapsto F(c_2)\times F(d_2)$, again mute in its contravariant variable, so by [L2] the inner end is the limit over the walking arrow of $d\mapsto F(c_2)\times F(d)$, which by step 1.1 is the value at $0$, namely $F(c_2)\times F(0)$. That chosen family is itself mute in $c_1$, so the outer end is the limit of $c\mapsto F(c)\times F(0)$, which is $F(0)\times F(0)$. Exchanging the roles of $\mathcal{C}$ and $\mathcal{D}$ runs the same two computations in the other order and gives $F(0)\times F(0)$ again, since the integrand is symmetric in the two pairs of variables.

The three objects therefore agree elementwise, and the identification is the identity of $F(0)\times F(0)$: an element $(z,w)$ has wedge component at $(c,d)$ equal to $(F_{0\to c}(z),F_{0\to d}(w))$, where $F_{0\to 0}$ is the identity and $F_{0\to 1}=F(u)$, and the same pair names the corresponding element of each iterated end. This is the conclusion [L1] predicts, computed here rather than quoted.

By [L1] the weighted limit is the set of natural transformations $W\Rightarrow F$. By [F5] such a transformation is a pair of functions ${\alpha }_0:\{p,q\}\to F(0)$ and ${\alpha }_1:\{\ast \}\to F(1)$ whose naturality equation at $u$ reads $F(u)\circ {\alpha }_0={\alpha }_1\circ W(u)$ as functions $\{p,q\}\to F(1)$; the right-hand side is constant at ${\alpha }_1(\ast )$, so the equation says $F(u)({\alpha }_0(p))={\alpha }_1(\ast )=F(u)({\alpha }_0(q))$.

So a natural transformation is exactly a pair $(s,t):=({\alpha }_0(p),{\alpha }_0(q))$ of elements of $F(0)$ with $F(u)(s)=F(u)(t)$, the remaining datum ${\alpha }_1(\ast )$ being determined as their common image. That set is the displayed pullback of $F(u)$ along itself, and by [F2] it carries the two projections and the universal property of a pullback.

The same answer comes from the category of elements. By [F6] the objects of $\int W$ are $(0,p)$, $(0,q)$ and $(1,\ast )$, and its non-identity morphisms are the two arising from $u$, one $(0,p)\to (1,\ast )$ and one $(0,q)\to (1,\ast )$, since $W(u)$ sends both $p$ and $q$ to $\ast$; so $\int W$ is a cospan. By [L2] the weighted limit is the limit of $F$ composed with the projection, which is the limit of $F(0)\to F(1)\leftarrow F(0)$, the pullback again.

For every set $X$, a function $f:X\to Y^S$ is exactly a family of functions $f_x:S\to Y$ indexed by $x\in X$, and hence exactly a function $\mathrm{\Phi }(f):S\to \mathbf{S}\mathbf{e}\mathbf{t}(X,Y)$ given by $\mathrm{\Phi }(f)(s)(x):=f(x)(s)$; conversely, from $\alpha :S\to \mathbf{S}\mathbf{e}\mathbf{t}(X,Y)$ one recovers $\mathrm{\Psi }(\alpha ):X\to Y^S$ by $\mathrm{\Psi }(\alpha )(x)(s):=\alpha (s)(x)$. These two constructions are inverse, so $\mathbf{S}\mathbf{e}\mathbf{t}(X,Y^S)\cong \mathbf{S}\mathbf{e}\mathbf{t}(S,\mathbf{S}\mathbf{e}\mathbf{t}(X,Y))$, which is the defining bijection of the power of $Y$ by $S$.

For every set $X$, a function $g:S\times Y\to X$ is exactly a function ${\mathrm{\Phi }}^{\mathrm{\prime }}(g):S\to \mathbf{S}\mathbf{e}\mathbf{t}(Y,X)$ given by ${\mathrm{\Phi }}^{\mathrm{\prime }}(g)(s)(y):=g(s,y)$; conversely, from $\beta :S\to \mathbf{S}\mathbf{e}\mathbf{t}(Y,X)$ one recovers ${\mathrm{\Psi }}^{\mathrm{\prime }}(\beta ):S\times Y\to X$ by ${\mathrm{\Psi }}^{\mathrm{\prime }}(\beta )(s,y):=\beta (s)(y)$. These two constructions are inverse, so $\mathbf{S}\mathbf{e}\mathbf{t}(S\times Y,X)\cong \mathbf{S}\mathbf{e}\mathbf{t}(S,\mathbf{S}\mathbf{e}\mathbf{t}(Y,X))$, which is the defining bijection of the copower of $Y$ by $S$.

Steps 1.1 and 1.2 identify the two objects explicitly, and [L1] says the same objects are, in general, the product and the coproduct of $S$ copies of $Y$. So in $\mathbf{S}\mathbf{e}\mathbf{t}$ the power is the function set and the copower is the Cartesian product.

The finite checks agree with those formulas: if $S=\{0,1\}$ and $Y=\{a,b,c\}$ then $Y^S$ has $3^2=9$ elements while $S\times Y$ has $2\cdot 3=6$; if $S=\{\ast \}$ then both objects identify with $Y$; if $S=\mathrm{\varnothing }$ then $Y^{\mathrm{\varnothing }}$ is a one-element set while $\mathrm{\varnothing }\times Y=\mathrm{\varnothing }$; and if $Y=\mathrm{\varnothing }$ then ${\mathrm{\varnothing }}^S=\mathrm{\varnothing }$ for nonempty $S$, but ${\mathrm{\varnothing }}^{\mathrm{\varnothing }}$ is again a one-element set.