# Homomorphisms, Tensor Products and Certain Canonical Maps.

Let and be vector spaces over a field and , their duals. In case is finite dimensional, one obtains a non-canonical isomorphism , a canonical isomorphism and a canonical isomorphism .

In case , and are not isomorphic: a basis of has a cardinality strictly larger than the one of . Moreover, the canonical map is still a monomorphism, but no longer surjective. In the case of , one has as well a canonical monomorphism , but it is no longer surjective as well. We want to study the images of the canonical maps and .

We begin with an auxiliary lemma.

Lemma.

Let with . Then there exists some with . Hence, if satisfies for all , then .

Proof.

Choose a -basis of such that there exists some with . Define by . Then and .

Proposition.

The map

is a monomorphism and its image is

In particular, if for some , then .

Proof.

Clearly, for , is -linear. Moreover, one quickly sees that is -linear itself. To see that is injective, let with . Now, by the lemma, there exists a with ; this shows that , whence . Therefore, an is injective.

Now, if , , whence . This shows that the image of is contained in the given set. Now assume that satisfies ; say, . Then for all , whence . By the Homomorphism Theorem, there exists a homomorphism such that

commutes. Now , whence . As (as ), is an isomorphism and we must have for some . But then, lies in the image of .

Finally, if , we saw that we have for any non-zero , with depending on . Since is injective, this shows that we must have .

This allows us to show that is surjective if, and only if, .

Corollary.

We have that is surjective if, and only if, .

Proof.

First, if , we see that . Since is injective, it follows that is in fact an isomorphism.

Now assume that . It suffices to construct a hyperplane in with ; this defines an element of which is not contained in the image of by the above proposition. For that, chose a basis of (using Zorn's lemma). This defines a family of elements of by , . Let be the subspace of generated by the 's. If we would have , we could emply Zorn's lemma a second time to obtain a hyperplane with ; this would prove our claim.

Hence, we have to show that . Note that for , ; in particular, for every , only finitely many of the 's are non-zero. Hence, it makes sense to define , . We claim that in case : for that, note that is a linear independent set in , since for every linear combination , we get for every .

Note that in fact, the proof shows that is isomorphic to a -fold direct product of , while is isomorphic to a -fold direct sum of . In case , these are of the same dimension, but in case , they are not.

We continue with the canonical map .

Proposition.

The map

is a monomorphism and its image is

the -vector space of finite dimensional -homomorphisms .

Proof.

One quickly sees that defines an element of , whence is well-defined and its image is contained in . Moreover, one quickly sees that is a homomorphism.

Let with , i.e. with for all . Without loss of generality, we can assume that our representation of satisfies that the 's are linearly independent. In that case, implies for all . But since this is true for all , it follows that for all . But then, . Therefore, , whence is injective.

Now let , and let be a basis of . Let be the projections with an for . Set . Then for all since for all ; therefore, . This shows that , whence we have equality.

Now is a -algebra, whence for , it makes sense to define . We are interested on how can be described in terms of and . This is resolved by the following result:

Proposition.

Let be -vector spaces. The map

is -linear and the following diagram commutes:

Proof.

Let , and be the canonical maps. Since these are isomorphisms, we have to show that for , and , we have . For that, let . Then

In particular, is a -algebra isomorphic to ; it posseses a if, and only if, .

Now consider transposition

Clearly, transposition is injective:

Lemma.

The map is -linear and injective.

Proof.

It is clear that is -linear. To see that it is injective, let with . Let ; then for all , whence by the first lemma. But that means .

We show that transposition restricts to the subspaces of the homomorphism spaces of homomorphisms with finite-dimensional image.

Lemma.

Let . The map

is an isomorphism. In particular, .

Proof.

Let . The map , is well-defined and a homomorphism as and . Now let with , i.e. with . But since is defined on , this means that . Hence, is injective.

Now let , i.e. there exists some with . Set ; then and . Therefore, is injective.

Finally, in case , we have and , whence . On the contrary, if , we have , whence .

Now we have seen that and in a canonical way, and we have the canonical monomorphism . We show that these maps behave nicely with transposition:

Proposition.

The map

is the unique homomorphism which makes the diagram

commuting.

Proof.

Let and . Then for , and

for all and . Hence, , what we had to show.

Now consider double transposition, i.e.

and its finite-dimensional image restriction

The above shows that using the canonical isomorphisms and , we can describe double transpotition by the following commuting diagram:

If , we obtain a map

Lemma.

The map is -linear and injective.

Proof.

First, if is fixed, for all , ; hence, . Now, if and , , we have

whence is -linear.

To see that is injective, let be such that . Let and ; since for all , we see that , but since this is the case for all we get .

Note that we have the following diagram:

Moreover, using the canonical embeddings and , we can define a map by , and a map by . It turns out that these map make the diagram commute:

Lemma.

The maps , and , are -linear and make the diagram

commute. In particular, is injective.

Proof.

That and are -linear is clear. For the lower triangle, let ; we have to show that . For that, let and ; then

For the right triangle, let ; we have to show that . For that, let and ; then

Now note that is injective. We can use this to determine the image of . For example, for ,

the last equivalence follows from the first proposition. Unfortunately, this criterion does not really helps in practice.

In case anyone knows a better description of the image of or , I'd be happy to know.