Checking a maps is good is a maps

Question

I want to prove that: \begin{align*} f:V\otimes \mathbb{C}&\rightarrow \mathbb{C}^{(I)}\\ \sum_{i\in I}e_i\otimes z_i &\mapsto (z_i) \quad\ \text{define a maps}\ \end{align*} where:
$\bullet$ $(e_i)$ is a basis of the $\mathbb{R}$- vector space $V$.
$\bullet$ the elements of $V\otimes \mathbb{C}$ have the form: $\sum_{i\in I}e_i\otimes z_i$ such that $(z_i)\in \mathbb{C}^{(I)}$
To show that $f$ is a maps, we have to mount two things:

1. $\forall x\in V \otimes \mathbb{C}\implies f(x)\in \mathbb{C}^{(I)}$
   (it is clear by construction)
2. $\forall x,y \in V \otimes \mathbb{C}\qquad x=y\implies f(x)=f(y) $
   I could not show condition 2)

Answer 1

It's never a good idea to define a homorphism directly on the tensor product, because you need to show that the map is well defined, after all tensors are just particular equivalence classes.

Instead, it's better to use the universal property of tensor product (UMP). Let's be clear. Let $V$ be an $\mathbb{R}$-vector space with base $\{e_i\}_{i\in I}$. $\mathbb{C}$ is also an $\mathbb{R}$-vector space. Let $$ h:V\times\mathbb{C}\rightarrow V\otimes_\mathbb{R}\mathbb{C}, $$ defined by $h(u,z)=u\otimes z$, for all $(u,z)\in V\times\mathbb{C}$. Define $g:V\times\mathbb{C}\rightarrow\mathbb{C}^{(I)}$, for all $(u,z)\in V\times\mathbb{C}$, with $u=\sum_{i\in I}a_ie_i,a_i\in\mathbb{R}$, $$ g(u,z)=g\left(\sum_{i\in I}a_ie_i,z\right)=(a_iz)_{i\in I}. $$ It is routine to check that $g$ is $\mathbb{R}$-bilinear, that is \begin{align} g(u+v,z)&=g(u,z)+g(v,z),\\ g(u,z+z')&=g(u,z)+g(u,z'),\\ g(au,z)&=ag(u,z)=g(u,az), \end{align} for all $u,v\in V,z,z'\in\mathbb{C},a\in\mathbb{R}$.

Now, by the UMP of the tensor product, there exists a unique homomorphism $f:V\otimes_\mathbb{R}\mathbb{C}\rightarrow \mathbb{C}^{(I)}$ making the following diagram commute (i.e. $f\circ h=g$) $\require{AMScd}$ \begin{CD} V\times\mathbb{C} @>{h}>> V\otimes_\mathbb{R}\mathbb{C}\\ @V g VV \\ \mathbb{C}^{(I)} \end{CD}

Of course, $f(u\otimes z)=f\left((\sum_{i\in I}a_ie_i)\otimes z\right)=(a_iz)_{i\in I}$.

Now, given $u=\sum_{i\in I}a_ie_i,v=\sum_{i\in I}b_ie_i\in V$, $a_i,b_i\in\mathbb{R}$, and $\omega,\omega'\in\mathbb{C}$ one has: \begin{align} u\otimes \omega+v\otimes\omega'&=(\sum_{i\in I}a_ie_i)\otimes\omega+(\sum_{i\in I}b_ie_i)\otimes\omega'\\&=\sum_{i\in I}e_i\otimes(a_i\omega)+\sum_{i\in I}e_i\otimes(b_i\omega')\\&=\sum_{i\in I}e_i\otimes(a_i\omega+b_i\omega')\\&=\sum_{i\in I}e_i\otimes z_i, \end{align} by setting $z_i=a_i\omega+b_i\omega'$. So every tensor can be written in this way.

Just like you observed, $f(\sum_{i\in I}e_i\otimes z_i)=(z_i)_{i\in I}$.