## Adjoint Isomorphism for Tensor and Hom (Pt. II)

**Point of Post: **This is a continuation of this post.

From all of this we can conclude that is a well-defined map (yay!). Let’s now verify that it is a group isomorphism. To see that is a homomorphism we merely note that

and since were arbitrary the desired equality holds. Lastly, let’s check that is a bijection. Injectivity is simple enough, if then we see that and agree on all simple tensors in and so (because the simple tensors are a generating set) must be equal on all of . For surjectivity, let be an arbitrary -map. Define the map given by . We claim that is an -biadditive -bimap. Indeed:

Thus, by the universal properties of the tensor product lifts to an -map such that . Note then that for all and so . Thus, surjectivity, and thus bijectivity, and thus isomorphism follow.

Let us fix and show that in is a natural transformation. Indeed, suppose that is an -map then we wish to show the commutativity of the following diagram

but this is easy:

and

The other naturalities follow similarly.

The case for is basically the same as the case for , and so we admit it.

Ok, so getting a little more down to earth, the above basically says that if we take a commutative ring and and take three -modules (thought of, in the usual way, as -bimodules) and noting that and are both -maps in this context, that’s pretty easy to see

for example, we have the following corollary (which is the case we’ll most often use):

**Theorem: ***Let be a commutative unital ring and be unital -modules. Then, the above maps and give -isomorphisms*

**References:**

[1] Dummit, David Steven., and Richard M. Foote. *Abstract Algebra*. Hoboken, NJ: Wiley, 2004. Print.

[2] Rotman, Joseph J. *Advanced Modern Algebra*. Providence, RI: American Mathematical Society, 2010. Print.

[3] Blyth, T. S. *Module Theory.* Clarendon, 1990. Print.

[4] Lang, Serge. *Algebra*. Reading, MA: Addison-Wesley Pub., 1965. Print.

[5] Grillet, Pierre A. *Abstract Algebra*. New York: Springer, 2007. Print.

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