Given a field $k$, let $M,N$ vector spaces with dual vector spaces $M^*, N^*$ respectively.

The following map $\varphi: M^*\otimes N \rightarrow Mod_K(M,N)$ is defined:

$$\varphi(f\otimes y)(x) = f(x) y \quad (*)$$

It is well defined as the map $F:M^* \times N \rightarrow Mod_K(M,N)$ given by $(*)$ is bilinear. I want to prove it is a $k$-linear injection. I'm able to prove the map is $k$-linear, let $x=\sum_j (f_j \otimes n_j), y = \sum_k (f_k \otimes n_k) \in M^*\otimes N $ and $a,b \in k$

$$ \varphi(ax + by)(m) = \varphi(\sum_i c_i f_i \otimes n_i)(m) = \sum_j a (f_j \otimes n_j)(m) +\sum_k b (f_k \otimes n_k)(m) = a\varphi(x) + b\varphi(y) $$

Lets prove is $\text{Ker}(\varphi) = \{0\}$ which would give us injectivity as the map is $k$-linear. For that let $x \in \text{Ker}(\varphi)$ we have that $\varphi(x) = 0$ which means that

$$ \varphi(x)(1) = \sum_i 1\otimes n_i = \sum_i 1\otimes \sum_j \lambda_{ij}e_j = \sum_j 1\otimes (\sum_i \lambda_{ij})e_j = 0$$

With $\{e_j\}$ a basis of $M$. And therefore $\sum_i \lambda_{ij}=0$. Now I want to prove $x=\sum_if_i\otimes n_i =\sum_if_i\otimes\sum_j\lambda_{ij}e_j=0$ but there is something I am missing.


Let $f_1,\ldots,f_m$ be a basis of $M^*$ and $y_1,\ldots,y_n$ be a basis of $N$ such that $\sum_{ij}k_{ij}f_i\otimes y_j$ lies in the kernel. Then

$0=\phi(\sum_{ij}k_{ij}f_i\otimes y_j)(x) = \sum_{ij} k_{ij}\phi(f_i\otimes y_j)(x) =\sum_{ij}k_{ij}f_i(x)y_j$ for all $x$.

Since $y_1,\ldots,y_n$ form a basis of $N$, the coefficient of $y_j$ must be 0 for all $j$,

$\sum_i k_{ij}f_i(x)=0$.

But $f_1,\ldots,f_m$ is a basis of $M^*$ and so

$k_{ij}=0$ for all $i$.

Since this holds for all $j$, it follows that $k_{ij}=0$ for all $i,j$.


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