0
$\begingroup$

In the Lecture Notes in Algebraic Topology by Davis & Kirk, we consider the Kronecker pairing which is defined by

$$H^n(C;R) \times H_n(C;R) \to R \\ ([\varphi],[\alpha]) \mapsto \varphi(\alpha)$$

where $H^n(C;R)$ is the cohomology and $H_n(C;R)$ is the homology with coefficients over a ring $R$.

Then it is said

Deduce by taking the adjoints that the Kronecker pairing defines a map $$H^n(C;R) \to \operatorname{Hom}(H_n(C;R),R).$$

I assume that this map is defined by $[\varphi] \mapsto ([\alpha] \mapsto \varphi(\alpha))$.

My question: What exactly does "taking the adjoints" mean? Does it has something to do with adjoint functors? Or maybe with the Adjoint Property of Tensor Products which is the following result

Proposition 1.14 (Adjoint Property of Tensor Products) There is an isomorphism of $R$-modules $$\operatorname{Hom}_R(A \otimes_R B, C) \simeq \operatorname{Hom}_R(A, \operatorname{Hom}_R(B,C))$$ natural in $A$, $B$, $C$ given by $\phi \leftrightarrow (a \mapsto (b \mapsto \phi(a \otimes b) ) ).$

It might be obvious but I just don't see it. I am thankful for any explanation!

$\endgroup$
  • $\begingroup$ Yeah, that is precisely it. A bilinear map is actually a linear map from the tensor product. And then the adjoint property of tensor products immediately gives you the map as said. $\endgroup$ – MooS Mar 20 '18 at 13:09
  • 1
    $\begingroup$ Note that the adjoint property of the tensor product is a specific case of adjoint functors : it implies that the functor $\_\otimes_RB $ is the left adjoint of the functor $Hom_R(B,\_)$. $\endgroup$ – Arnaud D. Mar 20 '18 at 13:11
2
$\begingroup$

In this situation, "taking the adjoint" means the following two steps. (1) By the universal property of the tensor product, view the Kronecker pairing as a homomorphism $H^n\otimes H_n\to R$. (2) Apply the adjoint property that you quoted to convert this homomorphism to a homomorphism $H^n\to\text{Hom}(H_n,R)$. Strictly speaking, "taking the adjoint" seems to refer to (2), but apparently (1) is to be understood. I suppose the author viewed (1) as so evident that it didn't need to be mentioned separately.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.