Categories, Functors and Natural Transformations — Examples

For each of the five structured categories in [L2], define $U$ on objects by $U(A)=$ the carrier of $A$, and define $U(f)$ to be the same ordered-pair relation as the structure-preserving map $f$, now regarded only as a function.

The underlying function of the identity morphism of $A$ is $1_{U(A)}$, so $U(1_A)=1_{U(A)}$.

Composition in every category in [L2] is composition of the underlying functions. Hence $U(g\circ f)=U(g)\circ U(f)$.

Thus the underlying-set assignments from $\mathbf{Grp}$, $\mathbf{Ring}$, $\mathbf{Vect}_F$, $R\text{-}\mathbf{Mod}$, and $\mathbf{Top}$ to $\mathbf{Set}$ are functors. They forget structure but not the identity and composition laws.

For a function $f:X\to Y$, the composite $X\xrightarrow fY\to F(Y)$ extends uniquely by [L2] to a homomorphism $F(f):F(X)\to F(Y)$.

Construct $R^{(X)}$ explicitly, since [L3] says only what it means for a module to be free and does not build one: let $R^{(X)}$ be the set of functions $a:X\to R$ whose support $\operatorname{supp}(a)=\{x:a_x\ne0\}$ is finite, with pointwise addition and scalar multiplication. Both operations preserve finite support because $\operatorname{supp}(a+b)\subseteq\operatorname{supp}(a)\cup\operatorname{supp}(b)$ and $\operatorname{supp}(ra)\subseteq\operatorname{supp}(a)$, so $R^{(X)}$ is a left $R$-module, and the family $e_x$ with $e_x(x)=1$ and $e_x=0$ elsewhere is a basis: every $a$ is the finite sum $\sum_{x\in\operatorname{supp}(a)}a_xe_x$, and a vanishing finite combination has every coefficient zero by evaluating at each index. So $R^{(X)}$ is free in the sense of [L3]. Now $R^{(f)}$ sends $a$ to the family $y\mapsto\sum_{x\in f^{-1}(y)\cap\operatorname{supp}(a)}a_x$; the index set is finite because it lies in $\operatorname{supp}(a)$, which is what [L4] requires, whereas $f^{-1}(y)$ itself may be infinite. The result again has finite support, contained in $f[\operatorname{supp}(a)]$, and $R^{(f)}$ is additive and $R$-linear because each coefficient is a finite sum of the corresponding coefficients of $a$. On basis elements it sends $e_x$ to $e_{f(x)}$.

Both maps assigned to $1_X$ fix every generator. The uniqueness of the free extensions therefore gives $F(1_X)=1_{F(X)}$ and $R^{(1_X)}=1_{R^{(X)}}$.

For $X\xrightarrow fY\xrightarrow gZ$, the maps $F(gf)$ and $F(g)F(f)$ agree on every generator. The module maps $R^{(gf)}$ and $R^{(g)}R^{(f)}$ likewise send $e_x$ to $e_{g(f(x))}$; finite-sum reindexing gives the same equality in coefficient form.

Hence $X\mapsto F(X)$ and $X\mapsto R^{(X)}$, with the maps above, define functors $\mathbf{Set}\to\mathbf{Grp}$ and $\mathbf{Set}\to R\text{-}\mathbf{Mod}$.

By [L2], matrix multiplication and identity matrices make $\mathbf{Mat}_F$ a category. Define $K:\mathbf{Mat}_F\to\mathbf{FinVect}_F$ by $K(n)=F^n$ and by letting $K(A)$ be multiplication by the matrix $A$.

Identity and composition are preserved by [L2] and [L3], so $K$ is a functor.

For every $m,n$, the map $A\mapsto K(A)$ is the coordinate bijection from $m\times n$ matrices to linear maps $F^n\to F^m$. Thus $K$ is fully faithful.

Suppose an ordered basis has been supplied for each finite-dimensional vector space $V$. If its length is $n_V$, the coordinate map $F^{n_V}\to V$ is a specified isomorphism, including when $V=0$. Hence these choices split essential surjectivity.

The criterion in [L5] now makes $K$ an equivalence. Thus chosen bases turn arbitrary finite-dimensional spaces into coordinate models without asserting that the two categories are strictly identical.

Let $\mathbf 2$ have objects $0,1$, their identities, and one further arrow $a:0\to1$. A functor $F:\mathbf2\to\mathbf{Set}$ is exactly a function $F(a):F(0)\to F(1)$.

Given functions $f:X\to Y$ and $g:X'\to Y'$, a natural transformation between their corresponding functors consists of maps $u:X\to X'$ and $v:Y\to Y'$ satisfying $v f=g u$. Thus it is exactly a commuting square.

Vertical composition composes the two side maps of commuting squares. It preserves the square equation, and identity transformations give identity squares. Therefore $[\mathbf2,\mathbf{Set}]$ is precisely $\mathbf{Set}^{\to}$ as described.

Let $\mathcal J$ have objects $E,V$, identities, and two distinct arrows $s,t:E\to V$. A functor $Q:\mathcal J\to\mathbf{Set}$ consists of a set $Q(E)$ of edges, a set $Q(V)$ of vertices, and source and target maps $Q(s),Q(t):Q(E)\to Q(V)$. This is exactly a quiver.

A natural transformation $\alpha:Q\Rightarrow Q'$ consists of functions $\alpha_E:Q(E)\to Q'(E)$ and $\alpha_V:Q(V)\to Q'(V)$ with $\alpha_VQ(s)=Q'(s)\alpha_E$ and $\alpha_VQ(t)=Q'(t)\alpha_E$. These are precisely the incidence-preservation equations for a quiver homomorphism.

Since identities and composition are componentwise in the functor category, this identification respects identity quiver maps and their composites. Hence quivers and quiver homomorphisms form $[\mathcal J,\mathbf{Set}]$.

Take the points of $X$ as objects and define $\operatorname{Hom}(x,y)$ to be the endpoint-preserving homotopy classes of paths from $x$ to $y$. For $\alpha:x\to y$ and $\beta:y\to z$, put $[\beta]\circ[\alpha]=[\alpha*\beta]$, and take the constant path at $x$ as $1_x$.

If $H$ deforms $\alpha$ to $\alpha'$ and $K$ deforms $\beta$ to $\beta'$ rel endpoints, paste $H(2s,t)$ for $s\le1/2$ to $K(2s-1,t)$ for $s\ge1/2$. The endpoint conditions agree along the seam, so the finite closed-pasting argument in [L2] gives an endpoint-preserving homotopy $\alpha*\beta\simeq\alpha'*\beta'$. Thus composition is well defined on classes.

Composing a path $\alpha$ with the straight-line homotopies from $s$ to $\max(0,2s-1)$ and to $\min(2s,1)$ proves the left and right identity laws. If $r(s)=2s$ for $s\le1/2$ and $r(s)=2-2s$ for $s\ge1/2$, then $\alpha((1-t)r(s))$ contracts $\alpha*\bar\alpha$ rel endpoints; the analogous formula $\alpha(t+(1-t)(1-r(s)))$ contracts $\bar\alpha*\alpha$. Hence reversal supplies a two-sided inverse class.

For three composable paths, let $\delta:[0,3]\to X$ traverse them successively. The two bracketings are $\delta\circ p$ and $\delta\circ q$, where $p(s)=4s$ for $s\le1/2$ and $p(s)=2s+1$ for $s\ge1/2$, while $q(s)=2s$ for $s\le1/2$ and $q(s)=4s-1$ for $s\ge1/2$. The formula $\delta((1-t)p(s)+tq(s))$ is an endpoint-preserving homotopy, so composition is associative on classes.

Steps 2.1, 3.1, and 2.2 make $\Pi_1(X)$ a groupoid. Its automorphisms at $x$ are the based-loop classes, but $[\beta]\circ[\alpha]=[\alpha*\beta]=[\alpha][\beta]$ in the traversal-order product of [L4]. Thus the identity on loop classes identifies $\operatorname{Aut}_{\Pi_1(X)}(x)$ with $\pi_1(X,x)^{\mathrm{op}}$. The inversion map $[\alpha]\mapsto[\bar\alpha]$ is therefore a canonical group isomorphism $\operatorname{Aut}_{\Pi_1(X)}(x)\cong\pi_1(X,x)$.

A partial function $A\rightharpoonup B$ is a function from a subset of $A$ to $B$, with the usual partial composition. Let $L(A)=A\amalg\{*\}$ and extend a partial function by sending every undefined input and the new point to the new point. This defines $L:\mathbf{Par}\to\mathbf{Set}_*$.

Conversely, let $R(X,x_0)=X\setminus\{x_0\}$. A pointed map $f:(X,x_0)\to(Y,y_0)$ induces the partial function defined at $x\ne x_0$ exactly when $f(x)\ne y_0$, with value $f(x)$. This defines $R:\mathbf{Set}_*\to\mathbf{Par}$.

The empty set is the unique zero object of $\mathbf{Par}$: if $Z$ were terminal, the empty partial function and each everywhere-defined map $\{*\}\to Z$ would force $Z=\varnothing$. In $\mathbf{Set}_*$ every pointed singleton is a zero object, so the distinct objects $(\{0\},0)$ and $(\{1\},1)$ are both zero objects.

Direct inspection of domains shows that both assignments preserve identities and partial composition. The canonical bijection $A\to RLA$ and the pointed bijection $(X\setminus\{x_0\})\amalg\{*\}\to X$ that is inclusion on the first summand and sends $*$ to $x_0$ are natural. Hence $RL\cong1_{\mathbf{Par}}$ and $LR\cong1_{\mathbf{Set}_*}$.

Step 2.1 supplies the equivalence in [L1]. An isomorphism as in [L2] would biject objects and, together with its inverse, preserve and reflect the zero-object property, contradicting step 1.3. Thus these categories are equivalent but not isomorphic.

Define $T_X(Y)=X\times Y$ with the product topology. For a continuous $f:Y\to Z$, define $T_X(f)=1_X\times f$ by $(x,y)\mapsto(x,f(y))$.

Its coordinate maps are the first projection and $f$ after the second projection, so $T_X(f)$ is continuous by [L2].

Pointwise, $1_X\times1_Y=1_{X\times Y}$ and $(1_X\times g)(1_X\times f)=1_X\times(gf)$.

Thus $T_X$ sends every morphism of $\mathbf{Top}$ to a morphism and obeys the two functor equations. It is an endofunctor by [L3].

Let $\mathcal O(X)$ be the open subsets of $X$ ordered by inclusion, and let $\mathcal C(X)$ be the closed subsets ordered by reverse inclusion. For $f:X\to Y$, assign to either kind of subset its inverse image under $f$.

Inverse image is monotone for inclusion and for reverse inclusion, preserves identity functions, and satisfies $(gf)^{-1}=f^{-1}g^{-1}$. Thus $\mathcal O,\mathcal C:\mathbf{Top}^{\mathrm{op}}\to\mathbf{Poset}$ are functors.

Complementation $c_X:\mathcal O(X)\to\mathcal C(X)$ is monotone because $U\subseteq V$ implies $X\setminus U\supseteq X\setminus V$. It is its own order-isomorphism inverse.

For every continuous $f:X\to Y$ and open $U\subseteq Y$, the identity $X\setminus f^{-1}(U)=f^{-1}(Y\setminus U)$ says exactly that the complement square commutes.

The componentwise order isomorphisms of step 2.2 are natural by step 2.3. Hence complementation gives $\mathcal O\cong\mathcal C$ as functors $\mathbf{Top}^{\mathrm{op}}\to\mathbf{Poset}$.

Define $\mathcal P(X)$ to be the power set of $X$ and $\mathcal P(f)(S)=f[S]$. Direct images satisfy $1_X[S]=S$ and $(gf)[S]=g[f[S]]$, so $\mathcal P:\mathbf{Set}\to\mathbf{Set}$ is a functor.

Define $\eta_X:X\to\mathcal P(X)$ by $\eta_X(x)=\{x\}$.

For every $x\in X$, $(\mathcal P(f)\eta_X)(x)=f[\{x\}]=\{f(x)\}=(\eta_Yf)(x)$. Therefore $\mathcal P(f)\eta_X=\eta_Yf$.

The equality in step 2.1 is the naturality square for every function $f$. Hence the singleton maps are the components of a natural transformation $\eta:1_{\mathbf{Set}}\Rightarrow\mathcal P$.

Let $G^{\mathrm{op}}$ have the same set and identity as $G$, with $a\star b=ba$. A homomorphism $f:G\to H$ is also a homomorphism $G^{\mathrm{op}}\to H^{\mathrm{op}}$, since $f(a\star b)=f(b)f(a)=f(a)\star f(b)$. Thus $G\mapsto G^{\mathrm{op}}$ defines an endofunctor $O$ on $\mathbf{Grp}$.

Define $\nu_G:G\to G^{\mathrm{op}}$ by $\nu_G(a)=a^{-1}$. Then $\nu_G(ab)=b^{-1}a^{-1}=\nu_G(a)\star\nu_G(b)$, and $\nu_G$ is its own inverse as a set map, so it is a group isomorphism.

Every homomorphism preserves inverses, so for $f:G\to H$ one has $O(f)\nu_G(a)=f(a^{-1})=f(a)^{-1}=\nu_Hf(a)$. Hence the component square commutes.

The isomorphisms $\nu_G$ are natural by step 2.1. Therefore inversion defines a natural isomorphism $1_{\mathbf{Grp}}\cong O$.

Use the tagged union $B\amalg C=(B\times\{0\})\cup(C\times\{1\})$. Define $A\times(B\amalg C)\to(A\times B)\amalg(A\times C)$ by $(a,(b,0))\mapsto((a,b),0)$ and $(a,(c,1))\mapsto((a,c),1)$.

Define $A^{B\amalg C}\to A^B\times A^C$ by restricting a function to the two tagged summands, and define $(A\times B)^C\to A^C\times B^C$ by composing with the two projections.

Untagging in step 1.1, joining two functions on disjoint tagged summands, and pairing two functions pointwise are respective two-sided inverses. Hence all three displayed maps are bijections.

Applying functions to the named entries before or after any map in steps 1.1 and 1.2 produces the same tuple or function value. Precomposition behaves the same way in each exponent variable. Thus every naturality square commutes in all covariant and contravariant variables.

The three componentwise bijections are natural by step 2.2, and their inverses are automatically natural. They therefore give the distributive and exponential natural isomorphisms.

For a commutative ring $R$, put $U(R)=R^\times$. For $\varphi:R\to S$, restrict $\varphi$ to units; it is a group homomorphism because ring homomorphisms preserve products, identities, and inverses. Identity and composition are inherited, so $U$ is a functor to $\mathbf{Grp}$.

Put $G_n(R)=\operatorname{GL}_n(R)$ and apply $\varphi$ entrywise. From the product formula, $\varphi((AB)_{ik})=\sum_j\varphi(a_{ij})\varphi(b_{jk})$, so this assignment preserves matrix products and identities and carries an inverse matrix to an inverse matrix. Entrywise identity and composition make $G_n$ a functor to $\mathbf{Grp}$.

Multiplicativity and the unit result in [L4] make $\det_R:G_n(R)\to U(R)$ a group homomorphism.

Applying $\varphi$ to the finite Leibniz sum term by term gives $\det_S(G_n(\varphi)(A))=\varphi(\det_R(A))=U(\varphi)(\det_R(A))$.

Step 2.2 is the naturality square for every commutative-ring homomorphism. Hence $(\det_R)_R$ defines a natural transformation $G_n\Rightarrow U$.

A functor $F:BG\to\mathbf{Set}$ selects one set $X=F(*)$ and, for every $g\in G$, a function $F(g):X\to X$.

Conversely, a left action defines $F(*)=X$ and $F(g)(x)=g\cdot x$; its action axioms are precisely the two functor equations.

The functor equations say $F(e)=1_X$ and $F(gh)=F(g)F(h)$. With $g\cdot x=F(g)(x)$, these become $e\cdot x=x$ and $(gh)\cdot x=g\cdot(h\cdot x)$, exactly the left-action axioms.

The two constructions recover the same functions $F(g)$ and the same action operation. Therefore left $G$-actions on sets are exactly functors $BG\to\mathbf{Set}$.

Define $X\mathbin{//}G$ to have objects $x\in X$ and one arrow $(g,x):x\to g\cdot x$ for each $g\in G$. Put $(h,g\cdot x)\circ(g,x)=(hg,x)$ and $1_x=(e,x)$.

Associativity and the identity laws follow from the corresponding group laws and the action law. The inverse of $(g,x)$ is $(g^{-1},g\cdot x)$, so this category is a groupoid.

Objects $x,y$ are joined by an arrow exactly when $y=g\cdot x$ for some $g$, which is exactly membership in the same orbit. An automorphism of $x$ is an element $g$ with $g\cdot x=x$, exactly an element of $\operatorname{Stab}_G(x)$.

Consequently the connected components of $X\mathbin{//}G$ are the $G$-orbits, and $\operatorname{Aut}(x)=\operatorname{Stab}_G(x)$ with the same multiplication.

Concatenation respects path homotopy. Let $\alpha\simeq\alpha'$ rel endpoints by $H$, and let $\beta\simeq\beta'$ rel endpoints by $K$, with the terminal point of $\alpha$ equal to the initial point of $\beta$. Setting $G(s,t)=H(2s,t)$ for $s\le\tfrac12$ and $G(s,t)=K(2s-1,t)$ for $s\ge\tfrac12$ gives a map on the two closed sets $[0,\tfrac12]\times I$ and $[\tfrac12,1]\times I$, which cover $I\times I$ and meet where $H(1,t)$ and $K(0,t)$ are both the shared endpoint. Each piece is a composite of $H$ or $K$ with an affine map of $I\times I$, so [L4] makes $G$ continuous, and it is a path homotopy $\alpha*\beta\simeq\alpha'*\beta'$ rel endpoints.

Reparametrisation does not change the class. Let $\varphi:I\to I$ be continuous with $\varphi(0)=0$ and $\varphi(1)=1$, and let $\lambda$ be a path. Then $H(s,t)=\lambda\bigl((1-t)\varphi(s)+ts\bigr)$ is continuous, starts at $\lambda\circ\varphi$, ends at $\lambda$, and is constant at each endpoint, so $\lambda\circ\varphi\simeq\lambda$ rel endpoints. Both bracketings of a triple concatenation, and each concatenation of a path with a constant path at its own endpoint, differ from the path itself precisely by such a $\varphi$. Hence concatenation is associative and the constant paths act as identities, in both cases up to path homotopy rel endpoints.

A path cancels its reversal. For a path $\lambda$ from $x$ to $y$, put $H(s,t)=\lambda(2s(1-t))$ for $s\le\tfrac12$ and $H(s,t)=\lambda(2(1-s)(1-t))$ for $s\ge\tfrac12$. The two closed pieces agree at $s=\tfrac12$, where both give $\lambda(1-t)$, so [L4] makes $H$ continuous. At $t=0$ it is $\lambda*\bar\lambda$ and at $t=1$ it is the constant path at $x$, and $H(0,t)=H(1,t)=x$ throughout. Thus $\lambda*\bar\lambda\simeq c_x$ rel endpoints, and applying this to $\bar\lambda$ gives $\bar\lambda*\lambda\simeq c_y$.

Define $\gamma_\#:\pi_1(X,x_0)\to\pi_1(X,x_1)$ by $\gamma_\#([\alpha])=[\bar\gamma*\alpha*\gamma]$, bracketed as $(\bar\gamma*\alpha)*\gamma$. The path $\bar\gamma$ runs from $x_1$ to $x_0$ and $\gamma$ from $x_0$ to $x_1$, so this is a loop at $x_1$. If $\alpha\simeq\alpha'$ rel endpoints, step 1.1 applied twice gives $\bar\gamma*\alpha*\gamma\simeq\bar\gamma*\alpha'*\gamma$, so $\gamma_\#$ is independent of the representative.