This is related to the definition of the tensor product space $V\otimes W$ as a qoutient space $C(V\times W)/N$, where N is some nice space
Let $C(S)$ be the $\mathbb{K}$-free vector space on an arbitrary set $S$, the latter constitutes a basis in the former. Let $H$ be a subspace whithin $C(S)$. Then the canonical projection $$ \pi:C(S) \rightarrow C(S)/H $$ is a surjective linear map, provided the set $C(S)/H$ has vector space structure, i.e., there is an addition and multiplication with scalars defined on it. (assumed to be known to you.)
Question: Is the space $C(S)/H$ spanned by the image of $S$ through $\pi$ ? i.e., is the following true $$ {\rm span}\{\pi(x)\ |\ x \in S \} = C(S)/H $$
My Answer: $$ \sum_{i=1}^n a_i \pi(x_i) = \pi\left(\sum_{i=1}^n a_i x_i\right) \hspace{1 cm} {\rm where} \hspace{1 cm} x_i \in S, a_i \in \mathbb{K}, n \in \mathbb{N} $$ Read forwards implies: ${\rm span}\{\pi(x)\ |\ x \in S \} \subseteq \ C(S)/H$. Read backwards implies: ${\rm span}\{\pi(x)\ |\ x \in S \} \supseteq \ C(S)/H$.
I'd like to know what is wrong about it and if this true for any vector space $V$, with a basis {$e_i$}, and a subspace $H \subseteq V$