Passing maps across an adjunction

Written by GPT-6.1 Sol (OpenAI), Ultra reasoning effort, October 2026. Self-checked by the writing AI (GPT-6.1 Sol, Ultra). Original text: CC0.

A map out of a free object is determined by what it does before the free construction. A map into a space of functions is determined by evaluating those functions. An adjunction puts these two patterns in one language. Its correspondence must commute with changes of both the source and the target. The resulting unit and counit record the universal maps, and two short equations make the correspondence reversible.

We begin with universal arrows, then recover and recognize the correspondence, compare adjoints, and calculate its effect on functions and modules. Use Points, fibres and universal representations for represented Hom families and their unique compatible comparisons; use Natural transformations and composition of functors for natural lifting and the component criterion. Basic references are [Stacks, Adjoints], Equivalences and chosen representatives, and Locally nilpotent operators and coinduced duals.

Fix a universe \(\mathcal U\). Categories have ambient sets of objects and \(\mathcal U\)-small Hom sets, interpreted with the Hom encoding in Points, fibres and universal representations, Section 1. They need not be essentially small. Natural transformations and choices indexed by all objects live in the ambient setting; choose a larger universe when a functor category is required as a small object. We use ambient choice for representing objects. All rings are unital, all modules are unital, and the ground ring \(k\) in Section 5 is commutative.

1. A best arrow to a target

Let \(L:C\to D\). For \(Y\in D\), the comma category \(L\downarrow Y\) has objects \((X,a)\) with \(a:LX\to Y\). An arrow \((X,a)\to(X',a')\) is an arrow \(t:X\to X'\) satisfying \(a' L(t)=a\). Identities and composites obey this condition by the functor laws.

A terminal object \((RY,\varepsilon_Y)\) in this category says exactly that every \(a:LX\to Y\) factors uniquely as

The arrow \(\varepsilon_Y\) therefore represents all maps to \(Y\) whose source has the form \(LX\).

Theorem 1.1. The following conditions on \(L\) are equivalent:

1. There is a functor \(R:D\to C\) and bijections

   \[ \begin{gathered} \Phi_{X,Y}:\operatorname{Hom}_D(LX,Y)\\ \longrightarrow\operatorname{Hom}_C(X,RY) \end{gathered} \tag{1.2} \]

   natural in both variables.

2. Every presheaf \(X\mapsto\operatorname{Hom}_D(LX,Y)\) is representable.

3. Every \(L\downarrow Y\) has a terminal object.

Proof interface and specialization. Conditions 2 and 1 are exactly the complete represented-Hom-family construction in Points, fibres and universal representations, Section 4. Its construction includes the identity and composition laws for \(R\), both-variable naturality, and compatible uniqueness. The element category of the presheaf in condition 2 has objects \((X,a)\) and arrows \(t\) with \(a'L(t)=a\), so it is precisely \(L\downarrow Y\). The complete terminal-universal-element criterion in Solution sets and universal representations, Section 1 gives 2 if and only if 3. These arguments use the stated size conventions and ambient choices, and require no essential smallness. \(\square\)

For intuition, chosen terminal objects directly determine \(R\) on arrows. Given \(v:Y\to Y'\), terminality in \(L\downarrow Y'\) gives the unique \(R(v):RY\to RY'\) with

The identity and composite satisfy this same characterization, proving the functor laws. Thus choices at separate objects become a coherent functor because their arrow maps are uniquely forced.

We call (1.2) an adjunction, write \(L\dashv R\), and call \(L\) the left adjoint and \(R\) the right adjoint. The bifunctors have domain \(C^{\mathrm{op}}\times D\). In concrete terms, for \(u:X'\to X\), \(v:Y\to Y'\), and \(f:LX\to Y\), naturality is

An arbitrary collection of objectwise bijections need not satisfy (1.4). Applying the theorem to opposite categories gives the left-adjoint criterion for a fixed right adjoint, with initial universal arrows in place of terminal ones.

2. The two maps that reverse the correspondence

Given \(\Phi\), define its unit \(\eta:1_C\Rightarrow RL\) and counit \(\varepsilon:LR\Rightarrow1_D\) by

Proposition 2.1. These are natural transformations. For \(f:LX\to Y\) and \(b:X\to RY\), the full correspondence is

They satisfy both triangle identities:

Proof. In (1.4), take the map \(1_{LX}\) and postcompose by \(f\). This gives the first formula in (2.2). Apply naturality of the inverse bijection to \(1_{RY}\), precomposed by \(b\), for the second.

For \(u:X\to X'\), the two ways to transpose \(L(u)\) give

For \(v:Y\to Y'\), the two ways to inverse-transpose \(R(v)\) give

These are exactly the naturality squares of the unit and counit. Finally, inverse-transpose \(\eta_X=\Phi(1_{LX})\) using (2.2); the result is the first identity in (2.3). Transpose \(\varepsilon_Y=\Phi^{-1}(1_{RY})\) for the second. \(\square\)

Equations (2.2) also identify two useful Hom squares. The map induced by \(R\) on \(\operatorname{Hom}_D(Y,Y')\), followed by inverse transposition, is \(v\mapsto v\varepsilon_Y\). The map induced by \(L\) on \(\operatorname{Hom}_C(X,X')\), followed by transposition, is \(u\mapsto\eta_{X'}u\). Both are statements about the specified adjunction, including its specified maps.

The following diagram shows the two excursions in (2.3). Each row starts and ends at the same object, and its composite is the identity. The rows belong to different categories.

Figure 2.1. The unit followed by the appropriate counit returns the original object and every map carried by it. The editable diagram is the displayed formula itself; it asserts exact identities, with no geometric approximation.

Theorem 2.2. Conversely, let \(L:C\to D\), \(R:D\to C\), and natural transformations \(\eta:1_C\Rightarrow RL\), \(\varepsilon:LR\Rightarrow1_D\) satisfy (2.3). Formula (2.2) defines an adjunction. Recovering its unit and counit by (2.1) returns exactly the given transformations.

Proof interface with the general formulas. Retain the complete two-inverse calculation following (5.2) in Descent through replacement objects, Section 5. That calculation uses only naturality and the two triangle identities: substitute \(Q=L\), \(H=R\). Its localization hypotheses establish those identities earlier in that lesson; they are not used in this calculation once the identities are assumed. Explicitly, its two composites become

For completeness, naturality of the forward map in both variables follows from

for \(u:X'\to X\). The inverse family is natural because the forward family is a natural family of bijections; alternatively (2.5) gives it directly. At identity maps, (2.2) gives \(\eta_X\) and \(\varepsilon_Y\), proving the final assertion. \(\square\)

3. When an adjoint retains all maps

Theorem 3.1. For a specified adjunction \(L\dashv R\):

1. \(L\) is fully faithful if and only if its unit \(\eta\) is invertible.
2. \(R\) is fully faithful if and only if its counit \(\varepsilon\) is invertible.
3. The following are equivalent: \(L\) is an equivalence; \(R\) is an equivalence; both are fully faithful; both specified transformations are invertible. In this case \(L\) and \(R\) are quasi-inverse equivalences.

Proof. For \(X,X'\in C\), compose the Hom map induced by \(L\) with the adjunction bijection:

By (2.4), this composite sends \(t\) to \(\eta_{X'}t\). If \(L\) is fully faithful, postcomposition by \(\eta_{X'}\) is bijective for every \(X\), so \(\eta_{X'}\) is invertible by the complete incoming-probe criterion in Points, fibres and universal representations, Section 2. Conversely, an invertible \(\eta_{X'}\) makes this composite bijective; \(\Phi\) is bijective, so the first arrow in (3.1) is bijective.

For \(Y,Y'\in D\), the corresponding composite is

It sends \(v\) to \(v\varepsilon_Y\). The complete outgoing-probe criterion proves assertion 2, in both directions.

Thus both adjoints are fully faithful exactly when both transformations are invertible, and those transformations exhibit quasi-inverses. Conversely, suppose \(L\) is an equivalence. The complete equivalence criterion in Equivalences and chosen representatives, Theorem 1.1 makes \(L\) fully faithful and essentially surjective. Hence \(\eta\) is invertible. The first triangle identity makes \(\varepsilon_{LX}=L(\eta_X)^{-1}\) invertible. For arbitrary \(Y\), choose an isomorphism \(t:LX\to Y\). Naturality gives

so \(\varepsilon_Y\) is invertible too. The argument with opposite categories treats an equivalence \(R\). This proves all implications, including empty categories. \(\square\)

Any specified quasi-inverse \(G\) of an equivalence \(F\) can serve as both its left and its right adjoint. Use the complete coherent-comparison construction in Equivalences and chosen representatives, Section 2 to obtain \(\alpha:GF\Rightarrow1\), \(\beta:FG\Rightarrow1\) with \(F\alpha=\beta F\) and \(\alpha G=G\beta\). The pair \((\alpha^{-1},\beta)\) gives \(F\dashv G\) by Theorem 2.2. The pair \((\beta^{-1},\alpha)\) gives \(G\dashv F\). No compatibility of initially arbitrary quasi-inverse comparisons is assumed.

Proposition 3.2. Suppose \(F:C\to D\) has adjoints \(A\dashv F\dashv B\). Then \(A\) is fully faithful if and only if \(B\) is fully faithful.

Proof. Write \(a\colon1_D\Rightarrow FA\), \(b\colon AF\Rightarrow1_C\) for the first unit and counit, and \(c\colon1_C\Rightarrow BF\), \(d\colon FB\Rightarrow1_D\) for the second. Successive transpositions give a bijection

Naturality of \(a\), followed by the first triangle identity for \(A\dashv F\), gives the exact square identity

If \(A\) is fully faithful, each \(a_X\) is invertible by Theorem 3.1. In (3.5), \(T\) and precomposition by \(a_X\) are bijections, so postcomposition by \(d_Y\) is bijective for all \(X\). Incoming probes make \(d_Y\) invertible; Theorem 3.1 makes \(B\) fully faithful. If \(B\) is fully faithful, postcomposition by \(d_Y\) is bijective. The same identity and bijectivity of \(T\) show that precomposition by \(a_X\) is bijective for every \(Y\). Outgoing probes make \(a_X\) invertible, hence \(A\) fully faithful. \(\square\)

The two outer adjoints need not be isomorphic. Exercise 1 gives a three-element ordered category where both are fully faithful but choose different objects.

There is also a recognition test that begins with a proposed universal arrow.

Proposition 3.3. Let \(F:C\to D\) be fully faithful, \(G:D\to C\), and \(a\colon1_D\Rightarrow FG\) natural. Suppose \(aF\) and \(Ga\) are invertible. Then \(G\dashv F\), with unit \(a\) and an invertible counit \(b\colon GF\Rightarrow1_C\).

Proof. The complete fully faithful postcomposition-lifting result in Natural transformations, Theorem 2.2 uniquely lifts \((aF)^{-1}\) to \(b\), with

It is a natural isomorphism, since its image is componentwise invertible and a fully faithful functor reflects isomorphisms. Equation (3.6) is the first triangle \(F(b_X)a_{FX}=1_{FX}\).

The other triangle does not follow just by cancelling \(a_Y\), which need not be invertible. Instead, naturality of \(a\) for \(a_Y\) gives

Apply \(G\), and cancel the invertible \(G(a_Y)\) on the right. The result is \(GFG(a_Y)=G(a_{FGY})\). Applying \(G\) to the first triangle at \(X=GY\) now gives

The natural isomorphism \(b\colon GF\Rightarrow1_C\) makes \(GF\) faithful: conjugating \(GF(t)\) by the components of \(b\) recovers \(t\). Therefore (3.8) implies \(b_{GY}G(a_Y)=1_{GY}\), the second triangle. Theorem 2.2 finishes the proof. \(\square\)

4. Compose correspondences and compare adjoints

Let \(L:C\to D\), \(R:D\to C\), \(L':D\to E\), \(R':E\to D\), with specified adjunctions \(L\dashv R\) and \(L'\dashv R'\). Composing their natural bijections gives

Thus \(L'L\dashv RR'\). Each step respects a map in either variable, so the composite does too. If \(\eta',\varepsilon'\) denote the second pair of maps, transposing the identities in (4.1), using (2.2), yields the precise composite maps

These are the unit and counit of the specified composite correspondence, so Proposition 2.1 proves their naturality and both triangles. For three adjunctions, both parenthesizations apply the same three Hom bijections in the same order. Hence the adjunction bijections, units and counits agree literally after the usual associative identification of functor composition.

Next fix adjunctions \(L_i\dashv R_i\) between \(C,D\), for \(i=1,2\). A transformation \(\alpha:L_1\Rightarrow L_2\) gives a transformation in the reverse direction, \(\alpha^*:R_2\Rightarrow R_1\), called its right mate.

Theorem 4.1. The two rules below are inverse bijections between natural transformations:

These transformations need not be invertible. Taking mates reverses vertical composition and takes identities to identities.

Proof. Precomposition by \(\alpha_X\) defines a map from \(\operatorname{Hom}_D(L_2X,Y)\) to \(\operatorname{Hom}_D(L_1X,Y)\), natural in \(X,Y\). Conjugating by the two adjunction bijections gives a natural map

The complete Yoneda lifting result in Points, fibres and universal representations, Section 2 gives a unique \(\beta_Y:R_2Y\to R_1Y\) inducing it. Naturality in \(Y\), followed by faithfulness of the same embedding, makes the \(\beta_Y\) natural. Evaluate (4.4) at \(X=R_2Y\), \(1_{R_2Y}\). Its inverse transpose is \(\varepsilon_{2,Y}\); precomposition gives \(\varepsilon_{2,Y}\alpha_{R_2Y}\); transposing by (2.2) gives the formula for \(\alpha^*\).

Conversely, postcomposition by any natural \(\beta:R_2\Rightarrow R_1\), conjugated back by the bijections, defines a map

natural in both variables. The covariant Yoneda lifting result in that same section uniquely represents it by an arrow \(\alpha_X:L_1X\to L_2X\). Naturality in \(X\) again gives a natural transformation. Evaluate at \(Y=L_2X\), \(1_{L_2X}\). Transposition gives \(\eta_{2,X}\), postcomposition gives \(\beta_{L_2X}\eta_{2,X}\), and inverse transposition gives the formula for \(\beta_*\).

The two constructions are inverse: conjugating a Hom map by bijections and conjugating back returns the same Hom map; the two fully faithful Yoneda embeddings make equality of those Hom maps equality of the represented transformations. This verifies both inverse assertions, rather than just one component formula. Composing two precomposition maps reverses the order of their representing arrows on the right. Faithfulness then gives \((\alpha'\alpha)^*=\alpha^*(\alpha')^*\); the identity Hom map is represented by the identity transformation. \(\square\)

The defining identity can also be written as a counit square:

Indeed \(\alpha^*_Y\) is the transpose under \(L_1\dashv R_1\) of the right side. This proves (4.6) directly and fixes every source and target. In the inverse formula (4.3), the last arrow has source \(L_1R_1L_2X\) and target \(L_2X\); it is a counit.

Corollary 4.2. A right adjoint, together with its specified adjunction, is unique up to a unique compatible natural isomorphism. The same holds for a left adjoint.

Proof interface. The complete compatible-uniqueness result for represented Hom families in Points, fibres and universal representations, Section 4 already proves the right-adjoint statement. With a fixed \(L=L_1=L_2\), Theorem 4.1 identifies its comparison \(R_2\to R_1\) with the mate of \(1_L\). The reverse comparison is its inverse, because mates reverse composition. The compatibility is (4.6); it characterizes the comparison uniquely by the adjunction and Yoneda. Opposite categories give the left-adjoint statement. \(\square\)

There may be several natural automorphisms of the bare functor. On abelian groups, \(1_{\mathsf{Ab}}\) has both the identity automorphism and the automorphism \(m\mapsto-m\); they differ on \(\mathbb Z\). Both are isomorphisms between the same functor and itself, but only the identity is compatible with two copies of the identity adjunction. One can also specify the adjunction whose Hom bijection sends \(f\) to \(-f\). Its unit and counit are both minus the identity, and (2.3) still holds. Specifying the correspondence matters.

5. Evaluation, scalar extension and coinduction

5.1 Functions of two arguments

Fix a set \(K\). The functors \(L_K(X)=X\times K\) and \(R_K(Z)=Z^K=\operatorname{Hom}_{\mathsf{Set}}(K,Z)\) act on arrows by product with \(1_K\) and by postcomposition. Identities and composites follow pointwise.

Currying and evaluation are inverse maps:

Both composites recover each value, proving bijectivity. Precompose by \(u:X'\to X\), postcompose by \(v:Z\to Z'\), or precompose the function argument by \(w:K'\to K\); evaluation at \(x',k'\) gives \(v(f(u(x'),w(k')))\) by either route. Thus the bijection is natural in \(X,Z\) and contravariantly compatible with a change of \(K\). In particular \(L_K\dashv R_K\).

Its unit sends \(x\) to the function \(k\mapsto(x,k)\). Its counit is evaluation \((h,k)\mapsto h(k)\). These formulas include \(K=\varnothing\), with the unique empty function, and all empty \(X,Z\). Exercise 2 checks the triangles and the fully faithful cases.

5.2 Tensoring over a central ground ring

Let \(R\) be a possibly noncommutative \(k\)-algebra, with \(k\) acting centrally, and let \(K\) be any \(k\)-module. Give \(N\otimes_k K\) the left \(R\)-action \(r(n\otimes t)=rn\otimes t\). Give \(\operatorname{Hom}_k(K,M)\) the left \(R\)-action \((rh)(t)=r h(t)\). Centrality makes this action preserve \(k\)-linearity. The action identities follow from those on \(M\). A module map in \(N\) tensors with \(1_K\), and a module map in \(M\) postcomposes; the universal property and pointwise composition give the two functor laws.

The complete balanced tensor and internal-Hom constructions in Operators, matrix relations and internal tensor–Hom, Sections 5 and 6, with their ordinary-module identifications in Hom, tensor and the exactness of module limits, Section 1, provide the underlying tensor and Hom objects. Currying gives

Here is the exact specialization of the balanced-map correspondence. A map \(f\) on the left gives \(g(n)(t)=f(n\otimes t)\); \(k\)-balance proves \(g(n)\) is \(k\)-linear, and \(f(rn\otimes t)=r f(n\otimes t)\) proves \(g\) is \(R\)-linear. Conversely, for \(g\) on the right, the map \((n,t)\mapsto g(n)(t)\) is biadditive and \(k\)-balanced, hence factors uniquely through \(N\otimes_k K\). Its \(R\)-linearity holds on pure tensors, which generate the tensor module. These rules are inverse on pure tensors and on each evaluated value. Precomposition, postcomposition, and a \(k\)-linear map \(K'\to K\) commute with the same evaluation formula. This is the adjunction \(-\otimes_k K\dashv\operatorname{Hom}_k(K,-)\) on left \(R\)-modules. No finite generation, projectivity or flatness of \(K\) is needed.

The unit sends \(n\) to \(t\mapsto n\otimes t\). The counit sends \(h\otimes t\) to \(h(t)\). These are the module versions of the two maps in Section 5.1.

5.3 Both adjoints of forgetting scalars

Let \(U:R\text{-}\mathsf{Mod}\to k\text{-}\mathsf{Mod}\) forget the \(R\)-action. Define \(E(K)=R\otimes_k K\). The left \(R\)-action is on the first factor. Restricting an \(R\)-linear map to \(1\otimes K\), and extending a \(k\)-linear map by multiplication, gives

The extension is balanced because the \(k\)-action is central and \(\lambda\) is \(k\)-linear. Restriction recovers \(\lambda\), and \(R\)-linearity recovers \(f\) on each \(r\otimes t\). Both rules commute with maps of \(K,M\), proving \(E\dashv U\). Its unit is \(t\mapsto1\otimes t\); its counit is multiplication \(r\otimes m\mapsto rm\).

For the other adjoint, define

The order \(ra\) is essential. It gives \(a(b\xi)(r)=\xi(rab)=((ab)\xi)(r)\), and the identity acts trivially. Postcomposition by a \(k\)-linear map in \(V\) gives an \(R\)-linear map and preserves functor composition.

The complete evaluation and inverse proof in Locally nilpotent operators and coinduced duals, Proposition 1.1 gives

Its proof applies to this arbitrary central ground ring and arbitrary target module, and verifies \(R\)-linearity, both inverses and both-variable naturality. Thus \(U\dashv D\). The unit at \(M\) is \(m\mapsto(r\mapsto rm)\); the counit at \(V\) is evaluation \(\xi\mapsto\xi(1)\).

We have \(E\dashv U\dashv D\). Proposition 3.2 compares full faithfulness of the two outer functors; it does not identify them. Exercise 3 makes the noncommutative order in \(D\) concrete.

6. Four exercises with full solutions

Exercise 1 — two different fully faithful adjoints

Introductory. Regard the chains \(C=\{0<1<2\}\) and \(D=\{0<1\}\) as categories. Let \(F:C\to D\) send \(0,1\) to \(0\), and \(2\) to \(1\). Find \(A\dashv F\dashv B\), list their units and counits, and decide which functors are fully faithful. Are \(A,B\) naturally isomorphic?

Solution. Define \(A(0)=0,A(1)=2\), and \(B(0)=1,B(1)=2\). These functions are monotone, hence define functors. For \(d=0\), \(A(d)\le c\) and \(d\le F(c)\) both hold for every \(c\). For \(d=1\), both hold exactly when \(c=2\). This proves \(A\dashv F\), since each Hom set in a poset is either empty or a singleton. For \(d=0\), \(F(c)\le d\) and \(c\le B(d)\) both hold exactly for \(c=0,1\). For \(d=1\), both hold for all \(c\). This proves \(F\dashv B\). Naturality is automatic here: all existing parallel arrows are equal.

For \(A\dashv F\), the unit \(d\to FA(d)\) is the identity at each \(d\). Its counit \(AF(c)\to c\) is respectively \(0\to0\), \(0\to1\), \(2\to2\). For \(F\dashv B\), the unit \(c\to BF(c)\) is \(0\to1\), \(1\to1\), \(2\to2\). Its counit \(FB(d)\to d\) is the identity at each \(d\). Both triangles are identities because their starting and ending objects agree and the categories are thin.

The maps \(A,B\) preserve and reflect order, so both are fully faithful. The functor \(F\) is faithful because its source Hom sets have at most one element. It is not full: \(\operatorname{Hom}_C(1,0)\) is empty, while \(\operatorname{Hom}_D(F1,F0)\) has one element. Finally, \(A(0)=0\) and \(B(0)=1\) are not isomorphic in the poset \(C\). Thus there is no natural isomorphism \(A\simeq B\), despite their common full faithfulness.

Exercise 2 — when cartesian adjoints lose no maps

Intermediate. For any set \(K\), verify both triangles for \(X\times K\dashv Z^K\). Determine exactly when its left adjoint, and exactly when its right adjoint, is fully faithful. Include \(K=\varnothing\).

Solution. The first triangle sends \((x,k)\) first to \((\eta_X(x),k)\), then evaluates the function \(\eta_X(x):t\mapsto(x,t)\) at \(k\). The output is \((x,k)\). The second triangle sends \(h:K\to Z\) to \(k\mapsto(h,k)\), then postcomposes this function with evaluation. At \(k\), the result is \(h(k)\), so the output is \(h\). For empty \(K\), the first equality is an equality of empty functions and the second an equality on the one-element function set. Neither requires a point of \(K\).

If the right adjoint is fully faithful, the counit at the singleton \(Z=1\) must be invertible. It is \(1^K\times K\to1\), canonically \(K\to1\), so \(K\) must have exactly one element. For such \(K\), evaluation is a bijection for every \(Z\), so the right adjoint is fully faithful.

For the left adjoint, the unit at \(X=\{0,1\}\) sends \(x\) to the constant function \(k\mapsto(x,k)\) in \((X\times K)^K\). If \(K=\varnothing\), its target has one element, so it is not injective. If \(K\) has two distinct elements \(k_0,k_1\), the function taking \(k_0\) to \((0,k_0)\), and every other \(k\) to \((1,k)\), lies outside the image; it is not one of the two constant-in-the-first-coordinate functions. Thus the unit is not surjective. If \(K\) is a singleton, every function \(K\to X\times K\) is uniquely of the required form, for every \(X\). Theorem 3.1 shows that the left adjoint is fully faithful exactly in this case too.

Exercise 3 — a matrix algebra checks the coinduced action

Advanced. Take \(k=\mathbb F_2\), \(R=M_2(k)\), and \(V=k\). Let \(E_{ij}\) be the matrix units and \(\xi_{ab}\in D(k)\) the coordinate functional with \(\xi_{ab}(E_{cd})=\delta_{ac}\delta_{bd}\). Compute the action of \(E_{ij}\) on \(\xi_{ab}\). Show concretely why replacing \(\xi(ra)\) by \(\xi(ar)\) in (5.4) fails as a left action. Finally verify both triangles for \(U\dashv D\) on elements.

Solution. Since \(E_{cd}E_{ij}=\delta_{di}E_{cj}\),

In particular \(E_{12}\xi_{22}=\xi_{21}\). The order is visible in the coordinate indices.

For the proposed wrong rule, write \(B_a\xi(r)=\xi(ar)\). Then

Set \(a=E_{12}\), \(b=E_{21}\), \(\xi=\xi_{11}\), \(r=E_{11}\). Since \(ba=E_{22}\) but \(ab=E_{11}\), we get

The left-action identity fails. For the correct rule \(A_a\xi(r)=\xi(ra)\), one instead has \(A_aA_b\xi(r)=\xi(rab)=A_{ab}\xi(r)\).

For any \(R\)-module \(M\), the unit sends \(m\) to \(h_m:r\mapsto rm\). Forget its action and apply the counit, evaluation at \(1\); the result is \(m\). For any \(V\) and \(\xi\in D(V)\), the unit at \(D(V)\) sends \(\xi\) to \(r\mapsto r\xi\). Apply \(D\) to the counit by postcomposing with evaluation at \(1\). At \(r\), the resulting function has value \((r\xi)(1)=\xi(r)\). Thus it returns \(\xi\), proving the other triangle. These computations apply to arbitrary \(k,R,V\), beyond the finite matrix example.

Exercise 4 — a transformation reverses its parameter map

Advanced. Let \(u:K\to H\) be a function. For the cartesian adjunctions, take \(\alpha_X=1_X\times u:X\times K\to X\times H\). Compute its right mate and recover \(\alpha\) with the inverse formula. For \(K=\{a,b\}\), \(H=\{c\}\), and \(Z=\{0,1\}\), decide whether the mate is invertible. Check how the mate behaves for a second parameter map \(v:H\to J\).

Solution. Here \(L_1=-\times K\), \(R_1=(-)^K\), \(L_2=-\times H\), \(R_2=(-)^H\). Apply the first formula in (4.3) to \(f:H\to Z\). The unit sends \(f\) to \(k\mapsto(f,k)\). Applying \(R_1(\alpha_{R_2Z})\) sends this to \(k\mapsto(f,u(k))\). Postcomposing with the second counit, evaluation, gives

For the inverse formula, start with \((x,k)\in X\times K\). Applying \(L_1(\eta_{2,X})\) gives \((h\mapsto(x,h),k)\). Applying \(L_1(\alpha^*_{L_2X})\) gives \((t\mapsto(x,u(t)),k)\). Evaluation with the first counit returns \((x,u(k))\). This recovers \(\alpha_X\) on every element, including the empty-domain cases.

For the stated finite sets, \(Z^H\) has two functions and \(Z^K\) has four. Precomposition sends the two functions to \((0,0)\) and \((1,1)\), respectively. It is injective and not surjective, so this mate is not invertible. The original \(\alpha_1:K\to H\), identified at \(X=1\), is surjective and not injective.

For \(v:H\to J\), the composite left transformation uses \(v u\). Its right mate sends \(f:J\to Z\) to \(f v u\), which is first the mate for \(v\) and then the mate for \(u\). This is exactly the reversal of vertical composition in Theorem 4.1, not a reversal of function composition inside the evaluated expression.

7. References

• [Stacks, Adjoints] The Stacks Project authors, Categories, “Adjoints.” The consulted structured source is the pinned categories chapter; its licence is GFDL 1.2 or later. The fully faithful criterion also has an official statement and proof. The linked source retains its licence; no source expression is incorporated into this lesson.
• [Universal representations] Points, fibres and universal representations and Solution sets and universal representations, complete proofs of the representation and universal-arrow interfaces used in Section 1.
• [Coherent comparisons] Natural transformations and composition of functors and Equivalences and chosen representatives, complete proofs of natural lifting and coherent quasi-inverses.
• [Module constructions] Operators, matrix relations and internal tensor–Hom, Hom, tensor and the exactness of module limits, and Locally nilpotent operators and coinduced duals, complete balanced-tensor and arbitrary-ground-ring coinduction interfaces.