# Proving this map is a k-linear injection

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 $$i,j$$.