`dustbringer.github.io`

Written by dustbringer on 14 March 2024 . View source.

Top\Hom as a (bi)module _B \Hom_{A_\ell}(_A M_B,_A K_C) _C _B \Hom_{C_r}(_A K_C,_B N_C) _A Tensor-Hom Adjunctions - \otimes_B N_C \dashv \Hom_{C_r}(_B N_C, -)_A M \otimes_B - \dashv \Hom_{A_\ell}(_A M_B, -)Remark

Tensor-Hom adjunctions for modules

To make the module structures explicit, we may write $_A M_B$ for an $(A,B)$ -bimodule $M$ . Also we write $\Hom_{R_\ell}(-,-)$ for morphisms of left $R$ -modules, and $\Hom_{R_r}(-,-)$ for morphisms of right $R$ -modules.

$\Hom$ as a (bi)module

Let $A,B,C$ be rings. Then let

$M$ be an $(A,B)$ -bimodule,
$N$ be a $(B,C)$ -bimodule,
and $K$ be an $(A,C)$ -bimodule.

$_B \Hom_{A_\ell}(_A M_B,_A K_C) _C$

The additive group $\Hom_{A_\ell}(_A M_B,_A K_C)$ can be endowed with the structure of a $(B,C)$ -bimodule. That is, for some left $A$ -module homomorphism $f: M \to K$ , define

the left $B$ -action to be $(b \cdot f)(m) = f(m \cdot b)$
and the right $C$ -action to be $(f \cdot c)(m) = f(m) \cdot c$ .

The left action is well-defined because $(b \cdot f)(a \cdot m) = f(a \cdot m \cdot b) = a \cdot f(m \cdot b) = a \cdot (b \cdot f)(m)$ , so $b \cdot f$ is a left $A$ -module homomorphism. Furthermore it forms a module because

(b_1 b_2 \cdot f)(m) = f(m \cdot b_1 \cdot b_2) = (b_2 \cdot f)(m \cdot b_1) = (b_1 \cdot (b_2 \cdot f))(m),

where the other properties follow trivially.

The right action is well-defined because $(f \cdot c)(a \cdot m) = f(a \cdot m) \cdot c = a \cdot f(m) \cdot c = a \cdot (f \cdot c)(m)$ , so $f \cdot a$ is a left $A$ -module homomorphism. It also forms a module since

(f \cdot c_1 c_2)(m) = f(m) \cdot c_1 \cdot c_2 = (f \cdot c_1)(m) \cdot c_2 = ((f \cdot c_1) \cdot c_2)(m).

This is a bimodule because $((b \cdot f) \cdot c)(m) = (b \cdot f)(m) \cdot c = f(m \cdot b) \cdot c = (f \cdot c)(m \cdot b) = (b \cdot (f \cdot c))(m)$ .

Notice that the bimodule structure was inherited from the "unused" right module structures of $M$ and $K$ .

$_B \Hom_{C_r}(_A K_C,_B N_C) _A$

Similarly on the right, we can endow the additive group $\Hom_{C_r}(_A K_C,_B N_C)$ with the structure of a $(B,A)$ -bimodule. That is, for some right $C$ -module homomorphism $f: K \to N$ , define

the left $B$ -action to be $(b \cdot f)(k) = b \cdot f(k)$
and the right $A$ -action to be $(f \cdot a)(k) = f(a \cdot k)$ .

The left action is well-defined because $(b \cdot f)(k \cdot c) = b \cdot f(k \cdot c) = b \cdot f(k) \cdot c = (b \cdot f)(k) \cdot c$ , so $b \cdot f$ is a right $C$ -module homomorphism. This has a module structure since

(b_1 b_2 \cdot f)(k) = b_1 \cdot b_2 \cdot f(k) = b_1 \cdot (b_2 \cdot f)(k) = (b_1 \cdot (b_2 \cdot f))(k).

The right action is well-defined because $(f \cdot a)(k \cdot c) = f(a \cdot k \cdot c) = f(a \cdot k) \cdot c = (f \cdot a)(k) \cdot c$ , so $f \cdot a$ is a right $C$ -module homomorphism. This gives a module structure with

(f \cdot a_1 a_2)(k) = f(a_1 \cdot a_2 \cdot k) = (f \cdot a_1)(a_2 \cdot k) = ((f \cdot a_1)\cdot a_2)(k).

This is a bimodule because $((b \cdot f) \cdot a)(k) = (b \cdot f) (a \cdot k) = b \cdot f (a \cdot k) = b \cdot (f \cdot a) (k) = (b \cdot (f \cdot a)) (k)$ .

Now notice that this is flipped (compared to above): the left $B$ -module structure was inherited from $_B N_C$ , and the right $A$ -module structure was inherited from $_A K_C$ . If we had it the same way around here as before, we could try and define an $(A,B)$ -bimodule structure by $(a \cdot f)(k) = f(a \cdot m)$ and $(f \cdot b)(k) = b \cdot f(k)$ . Again we can prove that $a \cdot f$ and $f \cdot b$ are right $C$ -module homomorphisms. However we run into problems when trying to prove it is a valid module structure. For the left action, we see that $(a_1 a_2 \cdot f)(k) = f(a_1 \cdot a_2 \cdot k) = (a_1 \cdot f)(a_2 \cdot k) = (a_2 \cdot (a_1 \cdot f))(k)$ is in the wrong order. For the right action, we see that $(f \cdot b_1 b_2)(k) = b_1 \cdot b_2 \cdot f(k) = b_1 \cdot (f \cdot b_2)(m) = ((f \cdot b_2) \cdot b_1)(m)$ also has the wrong order.

Tensor-Hom Adjunctions

Let $A,B,C,D$ be rings. Then let

$M$ be an $(A,B)$ -bimodule,
$N$ be a $(B,C)$ -bimodule,
$K$ be a $(D,C)$ -bimodule,
and $L$ be a $(A,D)$ -bimodule

$- \otimes_B N_C \dashv \Hom_{C_r}(_B N_C, -)$

We prove that $_D \Hom_{C_r}(_A M \otimes_B N_C, _D K_C)_A \simeq {_D\Hom_{B_r}(_A M_B, _D \Hom_{C_r}(_B N_C,_D K_C)_B)_A}$ as $(D,A)$ -bimodules.

Define $\phi: \Hom_{C_r}(M \otimes_B N, K) \to \Hom_{B_r}(M, \Hom_{C_r}(N,K))$ to map

\phi: f \mapsto (m \mapsto (n \mapsto f(m \otimes n))).

This is an $(D,A)$ -bimodule homomorphism because

$\phi(f + g) = (m \mapsto n \mapsto f(m \otimes n) + g(m \otimes n)) = \phi(f) + \phi(g)$
and

\begin{align*} \phi(d \cdot f \cdot a)(m)(n) &= (d \cdot f \cdot a)(m \otimes n) \\ &= d \cdot f(a \cdot m \otimes n) \\ &= d \cdot \phi(f)(a \cdot m)(n) \\ &= (d \cdot (\phi(f) \cdot a)(m))(n) \\ &= (d \cdot \phi(f) \cdot a)(m)(n). \end{align*}

Conversely, define $\psi: \Hom_{B_r}(M, \Hom_{C_r}(N,K)) \to \Hom_{C_r}(M \otimes_B N, K)$ to map

\psi: g \mapsto (m \otimes n \mapsto g(m)(n)).

This is an $(D,A)$ -bimodule homomorphism because

$\psi(f + g) = (m \otimes n \mapsto f(m)(n) + g(m)(n)) = \psi(f) + \psi(g)$
and

\begin{align*} \psi(d \cdot f \cdot a)(m \otimes n) &= (d \cdot f \cdot a)(m)(n) \\ &= (d \cdot f(a \cdot m))(n) \\ &= d \cdot f(a \cdot m)(n) \\ &= d \cdot \psi(f)(a \cdot m \otimes n) \\ &= (d \cdot \psi(f) \cdot a)(m \otimes n). \end{align*}

These functions are clearly inverses (by construction). Explicitly, this looks like

\begin{align*} \psi \circ \phi (f)(m \otimes n) &= \psi(m \mapsto n \mapsto f(m \otimes n))(m \otimes n) \\ &= (m \otimes n \mapsto (m \mapsto n \mapsto f(m \otimes n)))(m \otimes n) \\ &= f(m \otimes n) \end{align*}

and

\begin{align*} \phi \circ \psi (g)(m)(n) &= \phi(m \otimes n \mapsto g(m)(n))(m)(n) \\ &= (m \mapsto n \mapsto (m \otimes n \mapsto g(m)(n)))(m)(n) \\ &= g(m)(n). \end{align*}

Therefore $\phi$ and $\psi$ define the desired bijection.

$_A M \otimes_B - \dashv \Hom_{A_\ell}(_A M_B, -)$

We prove that $_C \Hom_{A_\ell}(_A M \otimes_B N_C, _A L_D)_D \simeq {_C \Hom_{B_\ell}(_B N_C, _B\Hom_{A_\ell}(_A M_B, _A L_D)_D)_D}$ as $(C,D)$ -bimodules.

Define $\phi: \Hom_{A_\ell}(M \otimes_B N, L) \to \Hom_{B_\ell}(N, \Hom_{A_\ell}(M, L))$ to map

\phi: f \mapsto (n \mapsto (m \mapsto f(m \otimes n))).

This is an $(C,D)$ -bimodule homomorphism because

$\phi(f + g) = (n \mapsto m \mapsto f(m \otimes n) + g(m \otimes n)) = \phi(f) + \phi(g)$
and

\begin{align*} \phi(c \cdot f \cdot d)(n)(m) &= (c \cdot f \cdot d)(m \otimes n) \\ &= f(m \otimes n \cdot c) \cdot d \\ &= \phi(f)(n \cdot c)(m) \cdot d \\ &= (\phi(f)(n \cdot c) \cdot d)(m) \\ &= (c \cdot \phi(f) \cdot d)(n)(m). \end{align*}

Conversely, define $\psi: \Hom_{B_\ell}(N, \Hom_{A_\ell}(M, L)) \to \Hom_{A_\ell}(M \otimes_B N, L)$ to map

\psi: g \mapsto (m \otimes n \mapsto g(n)(m)).

This is an $(C,D)$ -bimodule homomorphism because

$\psi(f + g) = (m \otimes n \mapsto f(n)(m) + g(n)(m)) = \psi(f) + \psi(g)$
and

\begin{align*} \psi(c \cdot f \cdot d)(m \otimes n) &= (c \cdot f \cdot d)(n)(m) \\ &= (f(n \cdot c) \cdot d)(m) \\ &= f(n \cdot c)(m) \cdot d \\ &= \psi(f)(m \otimes n \cdot c) \cdot d \\ &= (c \cdot \psi(f) \cdot d)(m \otimes n). \end{align*}

These functions are clearly inverses (by construction). The calculation looks very similar to the previous adjunction.

Remark

All the above isomorphisms also hold for left modules and right modules separately.

dustbringer 2026

Tensor-Hom adjunctions for modules

Hom⁡\HomHom as a (bi)module

BHom⁡Aℓ(AMB,AKC)C_B \Hom_{A_\ell}(_A M_B,_A K_C) _CB​HomAℓ​​(A​MB​,A​KC​)C​

BHom⁡Cr(AKC,BNC)A_B \Hom_{C_r}(_A K_C,_B N_C) _AB​HomCr​​(A​KC​,B​NC​)A​