Monoidal adjunctions, mates and dual projection formulas

Written by GPT-6.1 Sol (OpenAI), October 2026. Public domain (CC0).

A right adjoint to a strong monoidal functor has canonical tensor maps. These maps preserve multiplication and units, but need not be invertible. The projection formula is a different comparison: a dualizable factor makes it invertible even when the right adjoint itself is only lax monoidal.

Let \(F:\mathcal C\to\mathcal D\) be strong monoidal, with comparisons \(F_2:FX\otimes FX'\to F(X\otimes X')\) and \(F_0:I_{\mathcal D}\to F(I_{\mathcal C})\). Fix an adjunction \(F\dashv G\), with unit \(\eta\) and counit \(\epsilon\). Work in suitable fixed universes; neither category is assumed additive or braided. Use Passing maps across an adjunction, Section 2 for the full Hom bijection and both triangle identities, and Tensor duality, Section 2 for the two tensor-dual adjunctions. We retain canonical associators and unitors whenever parentheses are suppressed, as justified by Strict monoidal models.

1. The lax structure forced by the counit

For a map \(b:A\to GY\), its adjoint is \(b^\flat=\epsilon_YF(b)\). The inverse correspondence sends \(f:FA\to Y\) to \(G(f)\eta_A\). Define \[ \begin{gathered} \theta_{Y,Z}:GY\otimes GZ\longrightarrow G(Y\otimes Z),\\ \theta_0:I_{\mathcal C}\longrightarrow G(I_{\mathcal D}) \end{gathered} \tag{1.1} \] by their adjoints \[ \begin{gathered} \theta_{Y,Z}^{\flat} =(\epsilon_Y\otimes\epsilon_Z)F_2^{-1},\\ \theta_0^{\flat}=F_0^{-1}. \end{gathered} \tag{1.2} \] The types in the first line are \(F(GY\otimes GZ)\to Y\otimes Z\). The comparison \(F_2\) must be inverted there.

Theorem 1.1. These maps make \(G\) lax monoidal. They are the unique lax structure for which the counit is monoidal with respect to the composite lax structure on \(FG\).

Proof: naturality and associativity. Naturality of the counit and of \(F_2\) makes the first map in (1.2) natural in both objects. The natural adjunction correspondence therefore makes \(\theta\) natural. Its associativity equation is \[ \begin{gathered} G(a^{\mathcal D})\theta_{Y\otimes Z,W}\\ \cdot(\theta_{Y,Z}\otimes1)\\ =\theta_{Y,Z\otimes W}(1\otimes\theta_{Z,W})\\ \cdot a^{\mathcal C}. \end{gathered} \tag{1.3} \] Both arrows start at \((GY\otimes GZ)\otimes GW\). To check them, take adjoints. For the left composite, use naturality of \(\epsilon\), then twice the first identity (1.2). This gives the tensor of the three counits, followed by the target associator, after the inverse comparison from \(F((GY\otimes GZ)\otimes GW)\) to \((FGY\otimes FGZ)\otimes FGW\). For the right composite, the same substitutions give the tensor of the three counits after the comparison through \(FGY\otimes(FGZ\otimes FGW)\). Naturality of the associator and the associativity axiom for \(F_2\) equate these two maps. Injectivity of the adjunction bijection proves (1.3).

Proof: units and uniqueness. The left unit equation is \[ \begin{gathered}G(l^{\mathcal D}_Y)\theta_{I_{\mathcal D},Y}(\theta_0\otimes1)\\ =l^{\mathcal C}_{GY}.\end{gathered} \tag{1.4} \] After taking adjoints, substitute \(\epsilon_{I_{\mathcal D}}F(\theta_0)=F_0^{-1}\) and the first formula (1.2). Naturality moves the remaining counit past the target unitor. The left unit axiom of \(F\) identifies the resulting comparison with \(\epsilon_YF(l^{\mathcal C}_{GY})\), the adjoint of the right-hand side. This proves (1.4). The right unit proof uses the same substitutions and the right unit axiom of \(F\), yielding \(G(r_Y)\theta_{Y,I}(1\otimes\theta_0)=r_{GY}\). Thus all lax axioms hold. Finally the counit equations for any such structure require \[ \begin{gathered} \epsilon_{Y\otimes Z}F(\theta_{Y,Z})F_2\\ =\epsilon_Y\otimes\epsilon_Z,\\ \epsilon_I F(\theta_0)F_0=1_I. \end{gathered} \tag{1.5} \] Since the comparisons of \(F\) are invertible, these force exactly the adjoints (1.2). The bijection forces the maps themselves, proving uniqueness. \(\square\)

2. The unit, the monad and transported algebra objects

The lax structure on \(T=GF\) has \[ \begin{gathered} t_{X,X'}=G(F_2)\theta_{FX,FX'},\\ t_0=G(F_0)\theta_0. \end{gathered} \tag{2.1} \] The unit is a monoidal transformation: \[ \begin{gathered} \eta_{X\otimes X'}\\ =t_{X,X'}(\eta_X\otimes\eta_{X'}),\\ \eta_{I_{\mathcal C}}=t_0. \end{gathered} \tag{2.2} \] Proof. Take adjoints in the first equation. The adjoint of its left side is identity by the triangle identity. In the right side, naturality moves \(G(F_2)\) past the counit, and (1.2) replaces the adjoint of \(\theta\). Naturality of \(F_2\) then brings the two \(F(\eta)\)'s next to their counits. Both cancel by the other triangle identity, leaving identity again. For the unit equation the same computation gives \(F_0\epsilon_I F(\theta_0)=F_0F_0^{-1}=1\). This proves both equations. \(\square\)

Consequently the adjunction is monoidal with these structures on both functors and with its actual unit and counit, not with separately chosen maps. Its ordinary monad multiplication is \(m_X=G(\epsilon_{FX})\). The two monad unit equations are the adjunction triangles. Its associativity is naturality of \(\epsilon\) at \(\epsilon_{FX}:FGFX\to FX\), followed by \(G\). It is a monoidal monad: the multiplication's tensor equation is \[ \begin{gathered} m_{X\otimes X'}T(t_{X,X'})t_{TX,TX'}\\ =t_{X,X'}(m_X\otimes m_{X'}). \end{gathered} \tag{2.3} \] Indeed substitute (2.1). Naturality of \(\theta\) moves the two \(G(\epsilon_F)\)'s inside it. Equation (1.5) replaces their tensor by the composite through \(F(\theta)F_2\), and naturality of \(F_2\) gives precisely the left side. The multiplication's unit equation \(m_I T(t_0)t_0=t_0\) follows from the unit part of (1.5) in the same substitution. Together with (2.2) these are all the monoidal compatibility equations.

The lax structure also sends an algebra \((B,b,e)\) in \(\mathcal D\) to an algebra with multiplication and unit \[ \begin{gathered}G(b)\theta_{B,B},\\G(e)\theta_0.\end{gathered} \tag{2.4} \] A left \(B\)-action \(s:B\otimes N\to N\) becomes the left \(GB\)-action \(G(s)\theta_{B,N}\). To verify associativity, expand two successive actions and use naturality of \(\theta\) to move the inner \(G(s)\) through it. Equation (1.3) compares the two resulting tensor routes, and applying \(G\) to the original action equation equates their last maps. Equation (1.4) proves the action unit; the same calculation with \(N=B\) gives the algebra laws, including its right unit. Applying \(G\) to a structure-preserving map and using naturality of \(\theta\) gives its transported structure equation. Thus all algebra and module morphisms are carried along. No inverse to \(\theta\) or equivalence of these categories has been used or asserted.

3. Monoidal mates reverse the direction

Suppose \(F'\dashv G'\) is another such adjunction, and \(\alpha:F\to F'\) is a monoidal natural transformation; its components need not be invertible. Define its right mate \(\alpha^*:G'\to G\) by requiring \[ \begin{gathered}\epsilon_YF(\alpha^*_Y)\\ =\epsilon'_Y\alpha_{G'Y}.\end{gathered} \tag{3.1} \] Existence and uniqueness follow from the adjunction bijection. Explicitly, \[ \begin{gathered}\alpha^*_Y\\ =G(\epsilon'_Y)G(\alpha_{G'Y})\eta_{G'Y}.\end{gathered} \tag{3.2} \] Naturality of the three factors proves naturality of this mate.

Proposition 3.1. The mate is monoidal for the canonical lax structures, and taking mates reverses composition.

Proof: tensor and unit. To prove \[ \begin{gathered}\alpha^*_{Y\otimes Z}\theta'_{Y,Z}\\ =\theta_{Y,Z}(\alpha^*_Y\otimes\alpha^*_Z),\end{gathered} \tag{3.3} \] take adjoints under \(F\dashv G\). For the left side, use (3.1), then naturality of \(\alpha\) at \(\theta'\), and then (1.2) for \(F'\). The result is \[ \begin{gathered}(\epsilon'_Y\alpha_{G'Y}\otimes\\ \epsilon'_Z\alpha_{G'Z})F_2^{-1}.\end{gathered} \tag{3.4} \] Here monoidality of \(\alpha\) changes \((F'_2)^{-1}\alpha_{A\otimes B}\) into \((\alpha_A\otimes\alpha_B)F_2^{-1}\). Substitute (3.1) once in each tensor factor of (3.4), and use naturality of \(F_2\). This is the adjoint of the right side of (3.3). For the unit equation, the adjoint of \(\alpha^*_I\theta'_0\) reduces to \((F'_0)^{-1}\alpha_{I_{\mathcal C}}=F_0^{-1}\), since \(\alpha_I F_0=F'_0\). It is therefore \(\theta_0\), as required.

Proof: composition. For \(\beta:F'\to F''\), the adjoint of \(\alpha^*_Y\beta^*_Y\) is first \(\epsilon'_Y\alpha_{G'Y}F(\beta^*_Y)\), by (3.1). Naturality of \(\alpha\) moves \(F(\beta^*_Y)\) to \(F'(\beta^*_Y)\), and (3.1) for \(\beta\) gives \(\epsilon''_Y\beta_{G''Y}\alpha_{G''Y}\). This is the defining adjoint of \((\beta\alpha)^*_Y\). Uniqueness proves \((\beta\alpha)^*=\alpha^*\beta^*\); it also proves that the mate of identity is identity. \(\square\)

4. The actual projection comparisons

Define the left and right comparisons by \[ \begin{gathered} p_{X,Y}=\theta_{FX,Y}(\eta_X\otimes1),\\ q_{Y,X}=\theta_{Y,FX}(1\otimes\eta_X). \end{gathered} \tag{4.1} \] Their types are \(X\otimes GY\to G(FX\otimes Y)\) and \(GY\otimes X\to G(Y\otimes FX)\). Naturality follows from the naturality already proved. The adjoint of the first is \[ p_{X,Y}^{\flat}=(1_{FX}\otimes\epsilon_Y)F_2^{-1}, \tag{4.2} \] and the right formula has \(\epsilon_Y\otimes1_{FX}\) instead. For (4.2), insert (4.1) into (1.2), move \(F(\eta_X)\) through \(F_2^{-1}\) by naturality, and cancel \(\epsilon_{FX}F(\eta_X)=1_{FX}\). This identifies the comparison as a particular mate, fixing its direction and normalization.

These comparisons have the action coherence \[ \begin{gathered} G(a^{\mathcal D})G(F_2^{-1}\otimes1)\\ \cdot p_{X\otimes X',Y}\\ =p_{X,FX'\otimes Y}(1\otimes p_{X',Y})\\ \cdot a^{\mathcal C}. \end{gathered} \tag{4.3} \] Both sides start at \((X\otimes X')\otimes GY\). Substitute (4.1) into the right side. Naturality moves \(\eta_X\) past the inner \(p\); the lax associativity (1.3) combines the two \(\theta\)'s. Equation (2.2) combines \(\eta_X\otimes\eta_{X'}\) into \(\eta_{X\otimes X'}\), leaving the inverse \(F_2\) and associator displayed on the left. The unit equation is \[ \begin{gathered}G(l^{\mathcal D}_Y)G(F_0^{-1}\otimes1)\\ \cdot p_{I_{\mathcal C},Y}=l^{\mathcal C}_{GY}.\end{gathered} \tag{4.4} \] It follows by substituting \(\eta_I=G(F_0)\theta_0\) and using naturality and (1.4). Thus the maps have the full left-action unit and associativity compatibility, even before they are isomorphisms.

5. A left dual makes the left comparison invertible

Suppose \(X\) has a left dual \(D\), with evaluation \(e:D\otimes X\to I_{\mathcal C}\) and coevaluation \(c:I_{\mathcal C}\to X\otimes D\), satisfying both triangle identities. A strong monoidal functor carries this dual pair to \(FD,FX\), with \[ \begin{gathered} \widetilde e=F_0^{-1}F(e)F_2,\\ \widetilde c=F_2^{-1}F(c)F_0. \end{gathered} \tag{5.1} \] Indeed, expand each target triangle. Naturality of \(F_2\), its associativity equation and its unit equations identify that composite with \(F\) of the corresponding source triangle. The adjacent inverse comparisons cancel, including the \(F_0\)'s. The two remaining composites are \(F(1_X)\) and \(F(1_D)\), proving both target triangles.

Theorem 5.1. For every \(Y\), the actual map \(p_{X,Y}\) is an isomorphism.

Proof. For every \(Z\) there is a natural chain of bijections \[ \begin{gathered} \operatorname{Hom}_{\mathcal C}(Z,X\otimes GY) \\\simeq\operatorname{Hom}_{\mathcal C}(D\otimes Z,GY)\\ \simeq\operatorname{Hom}_{\mathcal D}(F(D\otimes Z),Y)\\ \simeq\operatorname{Hom}_{\mathcal D}(FD\otimes FZ,Y)\\ \simeq\operatorname{Hom}_{\mathcal D}(FZ,FX\otimes Y)\\ \simeq\operatorname{Hom}_{\mathcal C}(Z,G(FX\otimes Y)). \end{gathered} \tag{5.2} \] The first and fourth are the proved tensor-dual adjunctions for the indicated left duals. The second and fifth are the given adjunction, and the third is precomposition with \(F_2\). All are bijections without exactness assumptions.

The chain sends a map \(h:Z\to X\otimes GY\) to \(p_{X,Y}h\). To verify this, its first step sends \(h\) to \((e\otimes1)(1_D\otimes h)\); the second follows by \(\epsilon_YF(-)\). At the fourth step insert \(\widetilde c\) and the map \(\widetilde e\) inherited from this first step. Substitute (5.1). The dual triangle cancels the inserted evaluation and coevaluation on \(FX\). Naturality of the comparisons and counit leaves exactly \((1_{FX}\otimes\epsilon_Y)F_2^{-1}F(h)\), whose adjoint is \(p_{X,Y}h\) by (4.2). Thus postcomposition by \(p_{X,Y}\) is bijective on every Hom set. Taking \(Z\) to be its target gives a right inverse; injectivity for \(Z\) its source makes that right inverse a left inverse too. Hence \(p_{X,Y}\) is invertible. \(\square\)

For an explicit inverse, start at \(G(FX\otimes Y)\), insert \(c\) on the left, and use \[ \begin{gathered} G(FX\otimes Y)\\ \xrightarrow{c\otimes1}X\otimes D\otimes G(FX\otimes Y)\\ \xrightarrow{1\otimes p_{D,FX\otimes Y}}\\ X\otimes G(FD\otimes FX\otimes Y)\\ \xrightarrow{1\otimes G(\widetilde e\otimes1)}X\otimes GY. \end{gathered} \tag{5.3} \] This is the inverse chain (5.2) applied to identity on the target, so it is the inverse just proved. The map \(p_{D,FX\otimes Y}\) here is only the already-defined comparison; its invertibility is not assumed. A left dual of \(D\) is not required.

For the right comparison, reverse tensor order while keeping the categories and arrows: \(A\star B=B\otimes A\), with associator \((a_{C,B,A})^{-1}\) and exchanged unitors. Its pentagon and triangle are the original equations read in reversed tensor order. The same adjunction is monoidal with comparisons \(F_{2,B,A}\). A right dual of \(X\) in the original category is a left dual in this reversed structure. Applying Theorem 5.1 there proves that \(q_{Y,X}\) is an isomorphism whenever \(X\) has a right dual. No braiding is involved.

6. Four graded exercises with full solutions

Exercise 1 — Coinduction has pointwise multiplication (introductory)

For a field \(k\) and the cyclic group \(C_3=\{1,g,g^2\}\), let \(F\) forget a representation's group action. Construct its right adjoint, its canonical lax tensor maps, and the algebra obtained from the unit object \(k\).

Solution. Put \(G(V)=\operatorname{Map}(C_3,V)\) with \((g\cdot f)(h)=f(hg)\). Given a linear map \(b:F(M)\to V\), its transpose sends \(m\) to the function \(h\mapsto b(hm)\). It is equivariant since its value on \(gm\) at \(h\) is \(b(hgm)\). Evaluation at \(1\) is the inverse correspondence: for an equivariant \(a:M\to G(V)\), equivariance gives \(a(m)(h)=a(hm)(1)\). These two checks also prove naturality. The counit is evaluation at \(1\).

The map \(\theta(f\otimes f')\) is the function \(h\mapsto f(h)\otimes f'(h)\). It is equivariant by the diagonal action, and evaluation at \(1\) is the required tensor of counits, so uniqueness in Theorem 1.1 identifies it with the canonical lax map. Its unit sends a scalar to the constant function. Applying (2.4) to \(k\) gives the algebra \(k^3\) with coordinatewise multiplication and diagonal unit. The generator acts by \((a_0,a_1,a_2)\mapsto(a_1,a_2,a_0)\), by the chosen right-translation formula. This is an algebra automorphism and fixes the unit. \(\square\)

Exercise 2 — Invariants are lax, and the integral comparison can fail (intermediate)

Let \(F\) give an abelian group the trivial \(C_2\)-action, and let \(G\) take invariants of a \(\mathbb Z[C_2]\)-module. Compute the lax structure. For the sign representation \(S=\mathbb Z\), test \(\theta_{S,S}\) and \(p_{\mathbb Z/2,S}\), and explain why the dual hypothesis does not apply.

Solution. An equivariant map from a trivial group lands exactly in the invariants, giving the natural adjunction directly. The strong comparisons of \(F\) are identity on underlying tensors. The canonical lax map sends \(m\otimes n\) to the same tensor in \((Y\otimes Z)^{C_2}\), and its unit is identity on \(\mathbb Z\). This follows either by evaluating its transpose or by the invariant inclusions. For \(S\), invariance requires \(s=-s\), so \(GS=0\). But \(S\otimes S\simeq\mathbb Z\) has trivial action. Thus \(\theta_{S,S}:0\to\mathbb Z\) is not invertible.

Take \(X=\mathbb Z/2\). Its source \(X\otimes GS\) is zero. The action on \(FX\otimes S\simeq\mathbb Z/2\) is trivial because \(-1=1\) modulo two; the target invariants are \(\mathbb Z/2\). Hence the actual projection map is \(0\to\mathbb Z/2\), also not invertible. Finally if \(X\) had a left dual \(D\), every map \(D\otimes X\to\mathbb Z\) would be zero: for each \(d\), it gives a homomorphism from \(\mathbb Z/2\) to \(\mathbb Z\), which must vanish, and pure tensors generate. A zero evaluation cannot satisfy the triangle giving \(1_X\). Thus \(X\) has no such dual. The example distinguishes the unit comparison, the lax tensor comparison and the projection comparison. \(\square\)

Exercise 3 — Scalar extension has a larger projection formula (advanced)

For a homomorphism of commutative unital rings \(R\to A\), use \(F(X)=A\otimes_R X\) and restriction of scalars \(G\). Determine the canonical lax maps and prove the projection comparison is invertible for every \(R\)-module \(X\). Does this make \(G\) strong monoidal?

Solution. The adjunction is extension and restriction of scalars, with unit \(x\mapsto1\otimes x\) and counit \(a\otimes y\mapsto ay\); these maps are inverse under the usual bilinear correspondence. The strong comparison of \(F\) multiplies the two \(A\)-coefficients. Formula (1.2) therefore identifies \(\theta_{Y,Z}\) with the quotient \(GY\otimes_R GZ\to G(Y\otimes_A Z)\); its unit is \(R\to A\).

The projection map is \[ x\otimes y\longmapsto(1\otimes x)\otimes_A y. \] Its inverse sends \((a\otimes x)\otimes_A y\) to \(x\otimes ay\). For the \(R\)-balancing relation inside \(A\otimes_R X\), moving \(r\) from \(x\) to \(a\) gives the same element \(x\otimes\varphi(r)ay\). For the \(A\)-balancing relation in the outer tensor, multiplying \(a\) by \(b\) or \(y\) by \(b\) gives the same element \(x\otimes aby\). Thus the inverse is well-defined. Its first composite is identity, and its other composite sends a generator to \((1\otimes x)\otimes_A ay=(a\otimes x)\otimes_A y\), also identity. No projectivity condition on \(X\) occurs. Nevertheless for \(R=\mathbb Z\) and \(A=\mathbb Z/2\), the lax unit \(R\to A\) is not invertible, so \(G\) is not strong monoidal. The dual hypothesis of Theorem 5.1 is sufficient, not necessary for every individual adjunction. \(\square\)

Exercise 4 — Coinduction and an infinite-product obstruction (advanced)

Repeat Exercise 1 for the infinite cyclic group \(\mathbb Z\). Let \(X=\bigoplus_{n\in\mathbb Z}k e_n\) have trivial action and take \(Y=k\). Determine the projection comparison and show that it is injective but not surjective. Contrast the finite-group case without assuming that \(X\) is finite-dimensional.

Solution. The same adjunction formulas work for an arbitrary group: \(G(V)\) is the space of all functions, without a finite-support restriction. In general the projection comparison is \[ \begin{gathered}p_{X,Y}(x\otimes f)(h)\\=hx\otimes f(h),\end{gathered} \] because the adjunction unit sends \(x\) to \(h\mapsto hx\), and \(\theta\) is pointwise. For the trivial action this is the natural map from \(X\otimes_k k^{\mathbb Z}\) to \(X^{\mathbb Z}\). Each source element is a finite sum \(\sum_i x_i\otimes f_i\), so its resulting function has values in the finite-dimensional span of the \(x_i\)'s. The function \(n\mapsto e_n\) does not, and is absent from the image. To prove injectivity, rewrite any finite tensor sum using a basis of the finite-dimensional span of its first factors. If the resulting function is zero, independence of those factors makes every scalar coefficient function zero. The tensor was therefore zero.

For a finite group and any representation \(X\), the comparison is invertible without a dimension bound. At each group element decompose the finitely many values of a target function as finite sums of pure tensors \(x\otimes y\). For a term at \(h\), send it to \(h^{-1}x\otimes\delta_h(y)\), where \(\delta_h(y)\) is supported at \(h\) with value \(y\). This is a well-defined linear map, since it is induced by the linear maps on each finite direct-sum component; it is independent of the chosen pure decomposition. After applying \(p\), that term is zero at every other point and equals \(x\otimes y\) at \(h\). In the other composite, the term from \(p(x\otimes f)\) at \(h\) returns \(x\otimes\delta_h(f(h))\); the finite sum is \(x\otimes f\). This proves both inverse identities. For an infinite group that sum is not available for arbitrary functions, which is exactly the obstruction in the example. \(\square\)

7. Further reading and licence

• Johannes Flake, Robert Laugwitz and Sebastian Posur, Projection formulas and induced functors on centers of monoidal categories, Sections 2.2 and 3.1–3.3, for monoidal adjunctions and the dual projection criterion. The ordinary constructions, coherence checks and inverse maps used here are proved above; this lesson does not require or assert their later results on centers.
• Tensor duality and coherent inverses, Section 2, for the tensor-Hom bijections with their two inverse calculations; and Strict monoidal models and transport of algebra objects, Section 5, for the strong-functor transport comparison.

The exposition and four worked solutions are released under CC0-1.0. Linked works retain their own copyrights and licences.