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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

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

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