Adjunctions Units and Counits: Examples

Under [L1], the identity of $F(X)$ transposes to the generator inclusion, so ${\eta }_X(x)$ is the one-letter word $x$. The identity function of $U(G)$ extends uniquely to ${\epsilon }_G$, hence ${\epsilon }_G$ evaluates a word by multiplying its letters in $G$.

On a generator $x$, the composite ${\epsilon }_{F(X)}F({\eta }_X)$ first makes the one-letter word whose letter is the word $x$, then evaluates it to $x$. The two homomorphisms agree on all generators, so the first identity in [F2] holds.

For $g\in G$, the composite $U({\epsilon }_G){\eta }_{U(G)}$ forms the one-letter word $g$ and evaluates it to $g$. Thus the second identity in [F2] holds.

If $X=\mathrm{\varnothing }$, the first check is equality of the unique homomorphisms from the trivial free group. If $G$ is trivial, every evaluated word is its identity. Thus both boundary cases obey the same formulas.

By [L1], the unit is the standard basis inclusion $x\mapsto e_x$. The counit is the unique linear extension of $1_{U(V)}$, so [F1] gives ${\epsilon }_V(\sum a_v{e}_v)=\sum a_{v}v$.

On a basis vector $e_x$, the composite ${\epsilon }_{k^{(X)}}{k}^{({\eta }_X)}$ sends $e_x$ to $e_{e_x}$ and then to $e_x$. Linearity and [F1] show that it is the identity on all of $k^{(X)}$.

On $v\in V$, the composite $U({\epsilon }_V){\eta }_{U(V)}$ sends $v$ to $e_v$ and then to $v$. Hence both triangle identities in [F2] hold.

When $X=\mathrm{\varnothing }$, $k^{(X)}$ is the zero vector space and step 2.1 is the unique linear endomorphism of it. When $V=0$, the formula in step 1.1 is the zero map.

Simultaneous evaluation. For $a=(a_1,\dots ,a_n)\in k^n$ define ${\mathrm{ev}}_a\text{\hspace{0.17em}⁣}:A\to k$ by induction on $n$. For $n=0$, [F1] gives $A=k$ and ${\mathrm{ev}}_a=1_k$. For the step, [F1] gives $k[x_1,\dots ,x_n]=k[x_1,\dots ,x_{n-1}][x_n]$, so [F3] applied with $\phi ={\mathrm{ev}}_{(a_1,\dots ,a_{n-1})}$ and $s=a_n$ yields a unital ring homomorphism ${\mathrm{ev}}_a\text{\hspace{0.17em}⁣}:A\to k$ fixing $k$ and sending each $x_i$ to $a_i$; on a single indeterminate its formula is that of [F2]. Write $f(a):={\mathrm{ev}}_a(f)$. Being a ring homomorphism, ${\mathrm{ev}}_a$ satisfies $(f-g)(a)=f(a)-g(a)$ and $(rf)(a)=r(a)f(a)$.

The zero polynomial lies in $I(X)$. If $f,g\in I(X)$ and $r\in A$, then step 1.1 gives $(f-g)(a)=0-0=0$ and $(rf)(a)=r(a)\cdot 0=0$ for every $a\in X$, so [F5] makes $I(X)$ an ideal.

If $S\subseteq T$, every common zero of $T$ is a common zero of $S$, so $V(T)\subseteq V(S)$. If $X\subseteq Y$, every polynomial vanishing on $Y$ vanishes on $X$, so $I(Y)\subseteq I(X)$.

By the two displayed definitions, $X\subseteq V(S)$ means exactly that $f(a)=0$ for every $a\in X$ and $f\in S$, which means exactly that $S\subseteq I(X)$.

Regard subsets and ideals as inclusion preorders. Steps 2.2 and 2.3 give the contravariant arrow correspondence required by [F6], hence $V$ and $I$ form the claimed Galois connection.

Applying step 2.3 to $S$ and $V(S)$ gives $S\subseteq I(V(S))$, hence $V(I(V(S)))\subseteq V(S)$ by step 2.2; the reverse inclusion holds because every polynomial in $I(V(S))$ vanishes on $V(S)$. Thus $VIV=V$.

Dually, $X\subseteq V(I(X))$ and inclusion reversal give $I(V(I(X)))\subseteq I(X)$, while the opposite inclusion follows from the definition of $V(I(X))$. Thus $IVI=I$.

A path of length $n$ is a composable $n$-tuple of edges. Concatenation of paths is associative, and the length-zero path at a vertex is a two-sided identity. Thus these data form a category $P(Q)$ by [F1], including when $Q_0$ is empty.

Given a quiver map $q:Q\to U(\mathcal{C})$, define $\widehat{q}:P(Q)\to \mathcal{C}$ on vertices by $q$ and on a path by composing the images of its edges in order; send an empty path to the appropriate identity. Associativity in [F1] makes this a functor.

For a quiver map $f=(f_0,f_1):Q\to Q^{\mathrm{\prime }}$ define $P(f):P(Q)\to P(Q^{\mathrm{\prime }})$ by $f_0$ on objects and by $(e_1,\dots ,e_n)\mapsto (f_1{e}_1,\dots ,f_1{e}_n)$ on paths; the image tuple is composable because $f$ commutes with source and target. It sends empty paths to empty paths and concatenations to concatenations, so it is a functor, and $P(1_Q)=1_{P(Q)}$ and $P(gf)=P(g)P(f)$ hold componentwise. Thus $P$ is a functor.

Any functor extending $q$ must have the values in step 2.1: induction on path length using [F2] forces the empty path to an identity and each longer path to the composite of its edge images. Hence the extension is unique.

Both $\mathbf{Q}\mathbf{u}\mathbf{i}\mathbf{v}$ and $\mathbf{C}\mathbf{a}\mathbf{t}$ here consist of small quivers and small categories, so a quiver map is a pair of functions between sets and a functor is a pair of functions between sets; each morphism collection is a subset of a set of functions and hence a set, making both categories locally small as [L1] requires. Restriction to vertices and edges and the extension $q\mapsto \widehat{q}$ are inverse by steps 2.1 and 3.1, and they commute with the actions of step 2.2 and with postcomposition by functors, so the bijection $\mathbf{C}\mathbf{a}\mathbf{t}(P(Q),\mathcal{C})\cong \mathbf{Q}\mathbf{u}\mathbf{i}\mathbf{v}(Q,U(\mathcal{C}))$ is natural and [L1] yields $P⊣U$.

Every functor $T:I\mathcal{G}\to \mathcal{C}$ sends inverses to inverses, so [F1] and [F2] force every image morphism into $\mathrm{Core}(\mathcal{C})$. Keeping the same object and morphism functions gives a unique factor $\bar{T}:\mathcal{G}\to \mathrm{Core}(\mathcal{C})$.

If $K:\mathcal{C}\to \mathcal{D}$ is a functor, it sends isomorphisms to isomorphisms and therefore restricts to $\mathrm{Core}(K):\mathrm{Core}(\mathcal{C})\to \mathrm{Core}(\mathcal{D})$. Identities and composites restrict unchanged, so $\mathrm{Core}$ is a functor.

The factorization in step 1.1 and inclusion give inverse bijections $\mathbf{C}\mathbf{a}\mathbf{t}(I\mathcal{G},\mathcal{C})\cong \mathbf{G}\mathbf{p}\mathbf{d}(\mathcal{G},\mathrm{Core}(\mathcal{C}))$. Their definitions by restriction show naturality in both variables.

The categories of small categories and small groupoids are locally small, so [L1] applied to step 2.1 gives $I⊣\mathrm{Core}$. The construction also covers the empty groupoid and empty category.

The defining equations cut out a vector subspace of the function set $V^G$ from [F6]. Pointwise operations preserve the covariance equation, and finite unions preserve finite coset support.

The formula $(a\cdot f)(g)=f(a^{-1}g)$ preserves the covariance equation and finite support. The equations in [F1] show that it is a left $G$-action, and pointwise operations show that the action is linear. Hence ${\mathrm{Ind}}_H^G(V)$ is a $G$-representation.

For an $H$-equivariant linear map $u:V\to {\mathrm{Res}}_H^G(W)$, set $\widehat{u}(f):={\sum }_{gH\in G\mathrm{/}H}g\cdot u(f(g))$, where zero summands are omitted. If $g$ is replaced by $gh$, the summand becomes $gh\cdot u(h^{-1}\cdot f(g))=g\cdot u(f(g))$, so it depends only on the coset. The nonzero index set is finite, and [F8] makes the empty-support value zero.

Define $j_V:V\to {\mathrm{Res}}_H^G{\mathrm{Ind}}_H^G(V)$ by $j_V(v)(h)=h^{-1}\cdot v$ for $h\in H$ and $j_V(v)(g)=0$ for $g\mathrm{\notin }H$. The subgroup axioms make this well-defined with support in $H$, and direct calculation gives $j_V(a\cdot v)=a\cdot j_V(v)$ for $a\in H$; it is linear by [F3].

Linearity follows termwise from [F3]. Left multiplication bijects the relevant coset sets, so reindexing with [F9] gives $\widehat{u}(a\cdot f)=a\cdot \widehat{u}(f)$. Thus $\widehat{u}$ is a $G$-equivariant linear map, naturally in $V$ and $W$.

Every $f$ has the finite decomposition $f={\sum }_{gH}g\cdot j_V(f(g))$: at any $x\in G$, only the coset $xH$ contributes, and its contribution is $f(x)$. Therefore a $G$-map $T$ satisfies $\widehat{(T{j}_V)}(f)=T(f)$.

The function $j_V(v)$ is supported on the single coset $H$, and its value at $1$ is $v$, so $(\widehat{u})j_V=u$. Together with step 4.1, the assignments $u\mapsto \widehat{u}$ and $T\mapsto T{j}_V$ are inverse natural bijections, proving ${\mathrm{Ind}}_H^G⊣{\mathrm{Res}}_H^G$.

Let ${\eta }_2:2\to 2$ transpose $0$ and $1$, and let ${\eta }_X=1_X$ for every set $X\mathrm{\ne }2$. These are correctly typed components in the category [F2].

For $i:1\to 2$, the left side of [F1] is $i{\eta }_1$, which sends $0$ to $0$, while the right side is ${\eta }_{2}i$, which sends $0$ to $1$. Thus the naturality equation fails.

The family is not a natural transformation and therefore cannot be a unit by [F3]. Correct component types alone do not suffice.

Regard $C_2$ as a one-object category $\mathcal{C}$ by [F1] and set $F=G=1_{\mathcal{C}}$. Choose the identity element as the component of $\eta$ and $z$ as the component of $\epsilon$.

Both transformations are natural because $C_2$ is abelian, but each triangle composite is $z1=z\mathrm{\ne }1$. Hence both identities in [F2] fail.

Replacing $\epsilon$ by the identity element makes both composites equal to $1$, recovering the identity adjunction and isolating the failure in the wrong counit.

Apply [F4] to $r$ and $m$, viewing naturals as nonnegative integers, and write uniquely $r=qm+s$ with $0\le s<m$. Define $G(r)=q$.

If $mn\le r$, [F2] gives $r=mn+d$. Divide $d$ by $m$ as $d=tm+s^{\mathrm{\prime }}$ with $0\le s^{\mathrm{\prime }}<m$. Then $r=(n+t)m+s^{\mathrm{\prime }}$, so uniqueness in [F4] gives $q=n+t$ and $n\le q$.

Conversely, if $n\le q$, write $q=n+t$ by [F2]. Then $r=qm+s=mn+(tm+s)$, so [F2] gives $mn\le r$. Hence $F(n)\le r$ exactly when $n\le G(r)$.

The equivalence in steps 2.1 and 2.2 is the condition [F5], so $F⊣G$. Taking $r=mn$ gives quotient $n$ and remainder $0$, hence $G(F(n))=n$; the counit is $mG(r)\le r$.

The equality $GF=1$ gives $FGF=F$, while applying $G$ to the counit formula and using the quotient $q=G(r)$ gives $GFG=G$. When $m=1$, every remainder is $0$ and both maps are the identity; the assumption $m>0$ excludes division by zero.

Three nonstrict consequences of [F7], each obtained by adjoining the equality case to a strict statement, are used below. (a) $t\le u$ exactly when $t+v\le u+v$: if $t<u$ then $t+v<u+v$ by [F7] and if $t=u$ then $t+v=u+v$, so $t+v\le u+v$; applying the same with $-v$ recovers $t\le u$. (b) $t\le u$ exactly when $-u\le -t$: translating by $v=-t-u$ carries the first to the second by (a), and translating by $t+u$ carries it back. (c) If $t\le u<w$ then $t<w$: for $t<u$ this is transitivity of the strict order and for $t=u$ it is immediate.

An integer $d$ with $d<1$ satisfies $d\le 0$. The order of $\mathbb{Z}$ is total by [F5], so either $d\le 0$ or $0<d$. In the second case $d\ge 0$ and $d\mathrm{\ne }0$, so [F5] presents $d$ as the image of a unique natural $j$, with $j\mathrm{\ne }0$ because the embedding is injective and sends $0$ to $0$; then $0<j$, so [F6] with $m=0$ gives $\sigma (0)=1\le j$, and the embedding preserves order and $1$, so $1\le d$. That contradicts $d<1$ by trichotomy, leaving $d\le 0$.

$\lfloor J(m)\rfloor =m$ for every integer $m$. Indeed $m\le J(m)<m+1$ holds because $J(m)$ is $m$ read in $\mathbb{R}$ and $0<1$ there, so the uniqueness clause of [F4] identifies $\lfloor J(m)\rfloor$ with $m$.

For every integer $m$ and real $x$: $J(m)\le x$ exactly when $m\le \lfloor x\rfloor$. If $m\le \lfloor x\rfloor$ then, since $\lfloor x\rfloor \le x$ by [F4] and $J$ preserves order by [F5], transitivity gives $J(m)\le x$. Conversely suppose $J(m)\le x$. By [F4] also $x<\lfloor x\rfloor +1$, so $m<\lfloor x\rfloor +1$ by step 1.1(c), and translating by $-\lfloor x\rfloor$ with [F7] turns this into $m-\lfloor x\rfloor <1$. Now $m-\lfloor x\rfloor$ is an integer by [F5], so step 1.2 gives $m-\lfloor x\rfloor \le 0$, that is $m\le \lfloor x\rfloor$.

$\lceil J(m)\rceil =m$ for every integer $m$: $-m$ is an integer with $J(-m)=-J(m)$ by [F5], and step 1.3 gives $\lfloor J(-m)\rfloor =-m$, whence $\lceil J(m)\rceil =-\lfloor -J(m)\rfloor =-\lfloor J(-m)\rfloor =m$.

$J$ and $\lfloor -\rfloor$ are monotone, and $J⊣\lfloor -\rfloor$. That $J$ is monotone is part of [F5]. If $x\le y$ then $J(\lfloor x\rfloor )\le x\le y$ by [F4], so step 2.1 applied to the integer $\lfloor x\rfloor$ and the real $y$ gives $\lfloor x\rfloor \le \lfloor y\rfloor$; thus $\lfloor -\rfloor$ is monotone. The equivalence of step 2.1 is then exactly the condition in [F2] for $J⊣\lfloor -\rfloor$, whose unit is $m\le \lfloor J(m)\rfloor$ and whose counit is $J(\lfloor x\rfloor )\le x$.

For every integer $m$ and real $x$: $\lceil x\rceil \le m$ exactly when $x\le J(m)$. By step 1.1(b), $x\le J(m)$ exactly when $-J(m)\le -x$, and $-J(m)=J(-m)$ because $J$ preserves addition and $0$ by [F5]. Step 2.1, applied to the integer $-m$ and the real $-x$, converts $J(-m)\le -x$ into $-m\le \lfloor -x\rfloor$, and step 1.1(b) converts that into $-\lfloor -x\rfloor \le m$, which is $\lceil x\rceil \le m$.

$\lceil -\rceil$ is monotone and $\lceil -\rceil ⊣J$. If $x\le y$ then $-y\le -x$ by step 1.1(b), so $\lfloor -y\rfloor \le \lfloor -x\rfloor$ by step 3.1, and step 1.1(b) gives $-\lfloor -x\rfloor \le -\lfloor -y\rfloor$, that is $\lceil x\rceil \le \lceil y\rceil$. With $J$ monotone by [F5], the equivalence of step 3.2 is the condition in [F2] for $\lceil -\rceil ⊣J$, whose unit is $x\le J(\lceil x\rceil )$ and whose counit is $\lceil J(m)\rceil \le m$.

Taking $\mathcal{C}=(\mathbb{Z},\le )$ and $\mathcal{D}=(\mathbb{R},\le )$ as thin categories by [F1], with $M=J$ and $L=\lceil -\rceil$, $R=\lfloor -\rfloor$, steps 3.1 and 4.1 supply the two adjunctions $L⊣M$ and $M⊣R$ required by [F3]. Hence $\lceil -\rceil ⊣J⊣\lfloor -\rfloor$ is an adjoint triple.

The counit of $J⊣\lfloor -\rfloor$ and the unit of $\lceil -\rceil ⊣J$ are in general strict, while steps 1.3 and 2.2 make the other unit and counit equalities. For $x=\frac{1}{2}$: $[F4]$ applied to $0\le \frac{1}{2}<1$ gives $\lfloor x\rfloor =0$, so $J(\lfloor x\rfloor )=0<\frac{1}{2}$; and applied to $-1\le -\frac{1}{2}<0$ it gives $\lfloor -x\rfloor =-1$, so $\lceil x\rceil =1$ and $\frac{1}{2}<1=J(\lceil x\rceil )$.