# Proving certain map is a k-linear isomorphism between $M^*\otimes N$ and $Mod_k(M,N)$

In Eiichi Abe Hopf Algebras volume we find the following exercice:

Given a field $$k$$, let $$M,N$$ be $$k$$-vector spaces with dual $$k$$-vector spaces $$M^*$$, $$N^*$$ respectively. Define a map $$\varphi:M^*\otimes N \rightarrow \text{Mod}_k(M,N)$$ for $$f \in M^*, y\in N,x \in M$$ by $$\varphi(f\otimes y)(x) = f(x)y$$ Then $$\varphi$$ is a $$k$$-linear injection. Moreover $$\varphi$$ is a $$k$$-linear isomorphism if $$M$$ or $$N$$ is finite dimensional.

I want to prove:

1) $$\varphi$$ is $$k$$-linear

2) $$\varphi$$ is an injection

3) $$\varphi$$ is an isomorphism under the hypothesis

• It is important that this isomorphism doesn't necessarily hold when M, N are not k-vector spaces – Diego Asterio Sep 23 '19 at 15:20

$$\varphi$$ is given over the generators of $$M^* \otimes N$$ therefore it is defined over $$M^* \times N$$ as $$\psi(f,n) = \varphi(f \otimes n)$$. We know that an isomorphism between $$Hom_k(M^* \otimes N,T)$$ and $$B_k(M \times N,T)$$ exists so to prove $$\varphi$$ is $$k$$-linear we can prove $$\psi$$ is $$k$$-bilinear.

Let $$a,b \in k; f,g \in M^*; y,y' \in N$$. For all $$x \in M$$:

$$\psi(af + bg, y)(x) = (af+bg)(x)y = ((af)(x)+(bg)(x))y = (af)(x)y + (bg)(x)y = (a(f(x))y + (b(g(x))y = a(f(x)y) + b(g(x)y) = a\psi(f,y)(x) + b\psi(g,y)(x)$$

$$\psi(f,ay + by')(x) = f(x)(ay+by') = f(x)ay + f(x)by' = af(x)y + bf(x)y' = a(f(x)y) + b(f(x)y') = a\psi(f,y)(x) + b\psi(f,y')(x)$$

which gives bilinearity thus $$\varphi$$ is $$k$$-linear.

To prove $$\varphi$$ a linear morphism is injective we are going to look at its kernel. Let an element belong to $$ker(\varphi)$$ for every $$x \in M$$ its image over $$x$$ is:

$$\varphi(\sum_{i=1}^n g_i \otimes n_i)(x) = \sum_{i=1}^n g_i(x)n_i = 0$$

For every $$i \in \{1,...,n\}$$ as $$N$$ is a $$k$$-vector space with basis $$\{e_\lambda\}_{\lambda \in \Lambda}$$ we have that for every $$i \in \{1,...,n\}$$ there exists a finite subset $$\{\lambda_i\}\subseteq \{\lambda\}_{\lambda \in \Lambda}$$ such that $$n_i = \sum_{\lambda_i}\alpha_{\lambda_i} e_{\lambda_i}$$ let $$\{\lambda_1, ..., \lambda_m\} = \cup_{i=1}^n(\{\lambda_i\})$$ we can express the sum above as $$\sum_{i=1}^n g_i(x)\sum_{j=1}^m \alpha_{\lambda_ij} e_{\lambda_j}$$ and so linear independence of $$\{e_{\lambda_1}, ..., e_{\lambda_m}\}$$ gives us for every $$x \in N$$ and every $$j \in \{1,...,m\}$$ $$\sum_{i=1}^n g_i(x)\alpha_{\lambda_ij} = 0$$ As this happens for every $$x \in N$$ it is shown that for every $$j \in \{1,...,m\} \sum_{i=1}^n \alpha_{\lambda_ij}g_i = 0$$ if we now express every $$g_i$$ in terms of the basis of $$M^*$$ $${f_\gamma}_{\gamma \in \Gamma}$$ we get that $$g_i = \sum_{\gamma_i }\beta_{\gamma_i} f_{\gamma_i}$$. As it is known $$\forall i \in \{1,...,n\} \{\gamma_i\}$$ is a finite subset of $$\Gamma$$ thus $$\cup_i \{\gamma_i\}$$ is also a finite subset. Changing its notation we have that $$\cup_i\{\gamma_i\} = \{\gamma_1, ..., \gamma_p\}$$ $$0 = \sum_{i=1}^n \alpha_{\lambda_ij} \sum_{k=1}^p \beta_{\gamma_jik} f_{\gamma_k} = \sum_{i=1}^n \sum_{k=1}^p \alpha_{\lambda_ij} \beta_{\gamma_jik} f_{\gamma_k} = \sum_{k=1}^p (\sum_{i=1}^n \alpha_{\lambda_ij} \beta_{\gamma_jik})f_{\gamma_k}$$ as $$\{f_\gamma\}$$ is a basis of $$M^*$$ we have that for every $$j$$ and for every $$k$$ $$(\sum_{i=1}^n \alpha_{\lambda_ij} \beta_{\gamma_jik}) = 0$$. We now rewrite the element of the kernel in term of this subsets of the basis $$z= \sum_{i=1}^n g_i \otimes n_i = \sum_{i=1}^n g_i \otimes \sum_{j=1}^m \alpha_{\lambda_ij} e_j = \sum_{j=1}^m \sum_{i=1}^n g_i \otimes \alpha_{\lambda_ij} e_j = \sum_{j=1}^m \sum_{i=1}^n \alpha_{\lambda_ij} \sum_{k=1}^p \beta_{\gamma_jik} f_{\gamma_k} \otimes e_j = \sum_{j=1}^m \sum_{i=1}^n \sum_{k=1}^p \alpha_{\lambda_ij} \beta_{\gamma_jik} f_{\gamma_k} \otimes e_j = \sum_{j=1}^m \sum_{k=1}^p (\sum_{i=1}^n \alpha_{\lambda_ij} \beta_{\gamma_jik}) f_{\gamma_k} \otimes e_j = \sum_{j=1}^m \sum_{k=1}^p 0 f_{\gamma_k} \otimes e_j = \sum_{j=1}^m 0 \otimes e_j = 0$$

Therefore $$ker(\varphi) = 0$$ and $$\varphi$$ is a $$k$$-linear injection

Now lets prove it is an isomorphism if $$dim(M)<\infty$$ or $$dim(N)<\infty$$. Suppose that $$N$$ is a finite dimensional $$k$$ vector space. Then if $$f \in Hom_k(M,N)$$ for every $$x \in f$$ we have that $$f(x) = \sum_{i=1}^n a_i(x) e_i$$ lets prove that $$\{a_i\}_{i \in {1,...,n}} \subset M^*$$ $$f(x+y) = \sum_{i=1}^n a_i(x+y) e_i$$ and $$f(x+y) = f(x) + f(y) = \sum_{i=1}^n a_i(x) e_i + \sum_{i=1}^n a_i(y) e_i$$ thus for every i $$a_i(x+y) = a_i(x) + a_i(y)$$ and $$a_i \in M^*$$. Now that $${a_i} \subset M^*$$ we have that $$\varphi(\sum_i a_i \otimes e_i) = f$$ which gives surjectivity.

Suppose now that $$M$$ is finite dimensional. For every $$x \in M$$ we have that $$f(x) = f(\sum_j b_j(x)e_j) = \sum_j f(b_j(x)e_j) = \sum_j b_j(x)f(e_j)$$ Lets prove that $${b_j} \subset M^*$$. We have that $$f(x+y) = f(x) + f(y) = \sum_j (b_j(x) + b_j(y))f(e_j)$$ and also that $$f(x+y) = \sum_j (b_j(x + y))f(e_j)$$ so $${b_j} \subset M^*$$. With all of that said $$\varphi(\sum_j b_j \otimes f(e_j)) = f$$ and $$\varphi$$ is surjective.