Reflective Subcategories and the Adjoint Functor Theorems — Examples

The constant functor $R:\mathbf{S}\mathbf{e}\mathbf{t}\to \mathcal{A}$ at $1$ is left adjoint to $I$: for every set $X$, both ${\mathrm{Hom}}_{\mathcal{A}}(1,1)$ and ${\mathrm{Hom}}_{\mathbf{S}\mathbf{e}\mathbf{t}}(X,1)$ contain one map, and these bijections are natural in $X$; by the converse clause of [L3] they determine a unit and counit satisfying the triangle identities, hence an adjunction $R⊣I$, so [L1] makes $\mathcal{A}$ reflective.

The sole object $1$ is initial in $\mathcal{A}$, so it is the empty colimit there by [L2]. Its image $I(1)=1$ is not initial in $\mathbf{S}\mathbf{e}\mathbf{t}$, whose initial object is $\mathrm{\varnothing }$.

Therefore the included empty colimit is not an ambient colimit, so $I$ does not preserve even this colimit.

A monomorphism $f:H\to \mathbb{Z}$ in $\mathbf{A}\mathbf{b}$ is injective: the inclusion $\ker f\to H$ and the zero map $\ker f\to H$ have equal composites with $f$, so they are equal and $\ker f=0$. Its corestriction onto the image $H^{\mathrm{\prime }}=f[H]\le \mathbb{Z}$ is then an isomorphism $H\to H^{\mathrm{\prime }}$ with $f={\iota }_{H^{\mathrm{\prime }}}\circ \bar{f}$ and ${\iota }_{H^{\mathrm{\prime }}}=f\circ {\bar{f}}^{-1}$, so $f$ mutually factors with the inclusion of $H^{\mathrm{\prime }}$ and represents it. By [L1], exactly one $n\in \mathbb{N}$ has $H^{\mathrm{\prime }}=n\mathbb{Z}$, so [L2] gives precisely the displayed representatives.

The inclusion $n\mathbb{Z}\hookrightarrow \mathbb{Z}$ factors through $m\mathbb{Z}\hookrightarrow \mathbb{Z}$ exactly when $n\mathbb{Z}\subseteq m\mathbb{Z}$, which holds exactly when $m$ divides $n$. By [L2] and [L3], this is the subobject order.

At $n=0$ the subgroup is $\{0\}$ and factors through every subgroup, whereas at $n=1$ it is all of $\mathbb{Z}$ and every subgroup factors through it. These are respectively the least and greatest classes.

A monomorphism $m:A\to X$ in $\mathbf{S}\mathbf{e}\mathbf{t}$ is injective: if $m(a)=m(a^{\mathrm{\prime }})$, the two maps $g,h:\{\ast \}\to A$ with $g(\ast )=a$ and $h(\ast )=a^{\mathrm{\prime }}$ satisfy $m\circ g=m\circ h$, so $g=h$ and $a=a^{\mathrm{\prime }}$. Conversely an injection is monic by the same cancellation. For such an $m$, the corestriction $\bar{m}:A\to m[A]$ is a bijection and satisfies $m={\iota }_{m[A]}\circ \bar{m}$, while ${\iota }_{m[A]}=m\circ {\bar{m}}^{-1}$. Thus $m$ mutually factors with the inclusion of its image and represents that subset by [L1] and [L2].

If the inclusions of subsets $S,T\subseteq X$ mutually factor, their image sets in $X$ coincide, hence $S=T$. Conversely equal subsets give the same inclusion. Therefore taking the image and taking the inclusion are inverse assignments between subobjects and subsets.

The argument also applies to the unique injection $\mathrm{\varnothing }\hookrightarrow X$, so the empty subset supplies the least subobject rather than an exceptional case.

For each $a\in A$, let $S_a=\{b\in B:a\le g(b)\}$. It is nonempty: because $g$ preserves the empty meet, $g({\mathrm{\top }}_B)={\mathrm{\top }}_A$, so $a\le g({\mathrm{\top }}_B)$. Completeness [L3] therefore supplies $f(a)=\bigwedge S_a$.

If $a\le a^{\mathrm{\prime }}$, then $S_{a^{\mathrm{\prime }}}\subseteq S_a$, so $\bigwedge S_a\le \bigwedge S_{a^{\mathrm{\prime }}}$. Hence $f$ is monotone and therefore a functor under [L1].

If $a\le g(b)$, then $b\in S_a$, so $f(a)=\bigwedge S_a\le b$.

Conversely, since $g$ preserves the meet of $S_a$, one has $g(f(a))={\bigwedge }_{c\in S_a}g(c)$. Every $g(c)$ on the right lies above $a$, hence $a\le g(f(a))$. Thus $f(a)\le b$ implies $a\le g(f(a))\le g(b)$ by monotonicity.

Steps 2.2 and 2.3 give the equivalence in [L2] for every $a,b$, so $f⊣g$. The empty-meet case in step 1.1 is what prevents the defining set from being empty.

Apply [L1] to the two-element set $S$. The normal subgroups of the set-sized group $F(S)$ form a set, so the displayed family is the promised solution set.

Let $C=\mathbb{Z}\mathrm{/}2\mathbb{Z}$ and map both $x$ and $y$ to its nonidentity element. Freeness extends this function uniquely to a homomorphism $\widehat{f}:F(S)\to C$. Its kernel $N=\ker \widehat{f}$ is a normal subgroup of $F(S)$ by [L2], so $N$ indexes a member of the family in step 1.1. Since $N\subseteq \ker \widehat{f}$, [L3] gives a unique $\bar{f}:F(S)\mathrm{/}N\to C$ with $\widehat{f}=\bar{f}{q}_N$, and $\bar{f}$ is injective because $\bar{f}(gN)=0$ forces $g\in \ker \widehat{f}=N$. Hence the original map factors through the member indexed by $N$.

More generally, [L1] gives this kernel-quotient factorisation for every map from $S$ into an underlying group, which verifies the solution-set property rather than only listing the quotients.

Both $m$ and $n$ are injective, and an injection $f$ in $\mathbf{S}\mathbf{e}\mathbf{t}$ is monic: $f\circ g=f\circ h$ gives $f(g(x))=f(h(x))$ for every $x$, hence $g(x)=h(x)$ and $g=h$. They are not the same map, since their domains $\{0\}$ and $\{\ast \}$ are distinct.

Let $u:\{0\}\to \{\ast \}$ and $v:\{\ast \}\to \{0\}$ be the unique maps. Then $m=n\circ u$ and $n=m\circ v$, so the monomorphisms mutually factor; [L2] also identifies $u$ and $v$ as inverse isomorphisms.

By [L1], $[m]=[n]$ even though $m\mathrm{\ne }n$.

Put one morphism $x\to y$ exactly when $x\le y$ in the displayed order. Every hom-collection is therefore empty or a singleton, so the definable-class category is locally small by [L5].

Every morphism in a thin category is monic: any parallel arrows that can be composed with it are already equal. Hence each arrow $m_{\alpha }:\alpha \to \mathrm{\infty }$ represents a subobject of $\mathrm{\infty }$.

The arrows $m_{\alpha }$ and $m_{\beta }$ mutually factor exactly when both $\alpha \le \beta$ and $\beta \le \alpha$, hence exactly when $\alpha =\beta$. Thus distinct ordinals give distinct subobject classes.

Suppose some set $M$ of monomorphisms into $\mathrm{\infty }$ contained a representative of every subobject class. By step 2.2 the only monomorphism into $\mathrm{\infty }$ that mutually factors with $m_{\alpha }$ is $m_{\alpha }$ itself, so $M$ would have to contain $m_{\alpha }$ for every ordinal $\alpha$, and $\alpha \mapsto m_{\alpha }$ is injective. Sending each such member of $M$ back to its domain would then exhibit the ordinals as the image of a set, making them a set and contradicting [L3]. No such $M$ exists, so the category is not well-powered by [L4], despite being locally small.

By [L1] and [L2], $G_n$ has componentwise addition. Every $(0,\bar{a})$ is killed by $n$. Conversely, if a nonzero integer $k$ kills $(z,\bar{a})$, then $kz=0$ in $\mathbb{Z}$, so $z=0$. Hence $\mathrm{Tor}(G_n)=\{0\}\times (\mathbb{Z}\mathrm{/}n\mathbb{Z})$.

The first projection ${\pi }_1:G_n\to \mathbb{Z}$ is a homomorphism by [L2], and it is surjective because ${\pi }_1(z,0)=z$ for every $z\in \mathbb{Z}$. Its kernel is $\{(0,\bar{a})\}=\mathrm{Tor}(G_n)$ by step 1.1, so the induced map $G_n\mathrm{/}\mathrm{Tor}(G_n)\to \mathbb{Z}$, $(z,\bar{a})+\mathrm{Tor}(G_n)\mapsto z$, is a well-defined surjective homomorphism, and it is injective because $z=0$ forces $(z,\bar{a})\in \mathrm{Tor}(G_n)$. It is therefore an isomorphism $G_n\mathrm{/}\mathrm{Tor}(G_n)\cong \mathbb{Z}$.

By [L3] the reflector sends $G_n$ to $G_n\mathrm{/}\mathrm{Tor}(G_n)$ with unit the quotient map, and by the equivalence in [L4] that unit is a universal arrow from $G_n$ to the inclusion: every homomorphism $f:G_n\to H$ with $H$ torsion-free factors uniquely through it. Step 2.1 identifies the target of that quotient map with $\mathbb{Z}$. Thus the computed object has the reflection's universal property.

When $n=1$, $\mathbb{Z}\mathrm{/}n\mathbb{Z}$ is the trivial group, so step 1.1 gives the zero torsion subgroup and steps 2.1–3.1 remain valid without dividing by a nontrivial integer.

A morphism $x\to y$ is by construction a set-indexed family of functions on the set $\mathrm{dom}(x)\cup \mathrm{dom}(y)$, so the morphisms $x\to y$ form a subset of the product of the function sets $Y_{\alpha }^{X_{\alpha }}$ over that index set. That product is a set, so every hom-collection is a set and $\mathcal{S}$ is locally small.

Let $D$ be a small diagram in $\mathcal{S}$ and let $T$ be the union of the supports of its set of objects, itself a set. Form the limit coordinatewise in $\mathbf{S}\mathbf{e}\mathbf{t}$: at $\alpha \in T$ take the limit $L_{\alpha }$ of the diagram of coordinates, and at $\alpha \mathrm{\notin }T$ every coordinate is $\{\mathrm{\varnothing }\}$, so the diagram there is constant at a one-element set and its limit is a one-element set. The code with domain $\{\alpha \in T:L_{\alpha }\text{ is not a singleton}\}$ and value $L_{\alpha }$ there is an object of $\mathcal{S}$, and its cone legs are the coordinatewise limit projections, the unique map being taken outside $T$. A cone over $D$ in $\mathcal{S}$ is exactly a coordinatewise cone, and the mediating map is coordinatewise unique, so this is a limit of $D$. The empty diagram has $T=\mathrm{\varnothing }$ and gives the code with empty domain. Every small diagram therefore has a limit, so [L2] makes $\mathcal{S}$ complete.

Let $\mathcal{H}$ be any small set of objects of $\mathcal{S}$. The union $T={\bigcup }_{H\in \mathcal{H}}\mathrm{dom}(H)$ is a set of ordinals. Were every ordinal a member of $T$ the ordinals would be a set, contradicting [L3], so some ordinal lies outside $T$; let $\beta$ be the least one, which is definable from $T$ and involves no selection.

Let $x$ be the code with empty domain, so $X_{\alpha }=\{\mathrm{\varnothing }\}$ for every $\alpha$, and let $y$ be the code with domain $\{\beta \}$ and $y(\beta )=\{0,1\}$. The two families $f,g:x\to y$ that send the single element of $X_{\beta }$ to $0$ and to $1$ respectively, and take the unique map at every other coordinate, are distinct morphisms. For every $H\in \mathcal{H}$ and every $h:y\to H$, the coordinate $H_{\beta }$ is $\{\mathrm{\varnothing }\}$ because $\beta \mathrm{\notin }\mathrm{dom}(H)$, so $h_{\beta }\circ f_{\beta }=h_{\beta }\circ g_{\beta }$; the two composites agree at every other coordinate as well, so $h\circ f=h\circ g$.

Therefore $\mathcal{H}$ does not detect the pair $f\mathrm{\ne }g$ and is not coseparating by [L1]. Since the argument applies to every small $\mathcal{H}$, $\mathcal{S}$ has no small coseparating set.