A proof of about independent linear subspaces of a vector space Let $M_1,M_2$ be linear subspaces of a vector space $V$ and let $M_1\times M_2$ be the product vector vector space. If the linear map $T: M_1\times M_1 \to V$ defined by $T(x_1,x_2) = x_1+x_2$ is injective then the linear subspaces $M_1,M_2$ are said to be independent. I'd like to show that $M_1,M_2$ are independent if and only if $M_1\cap M_2 = \{\theta\}$. Here $\theta$ is the zero vector of $V$.
Proof.
$(\implies)$ I only have to show $M_1\cap M_2 \subseteq \{\theta\}$ since the other inclusion holds by definition. Let $x,y\in M_1\cap M_2$, we claim that $x=y$, thus it must be that $x=\theta$. By the commutativity axiom of V,  $$T(x,y) = x+y = y+x = T(y,x).$$ But $T$ is one-to-one, thus
$$(x,y) = (y,x),$$ so $x=y$, Q.E.D.
Do you agree with this proof?
Note: this question is similar to but not the same as Prove that subspaces $U,V$ are linearly independent iff $U \cap V = \{0\}$. The proofs/hints provided there were somehow assuming $V$ has a basis, which here is not the case. Furthermore, I'm working under a precise definition of independence of linear subsets (see above), which was not given that question.

Your proof for $\Rightarrow$ is correct (though a bit overcomplicated : it is simpler to take $y=0$ from the beginning). To show the converse, assume $M_1 \cap M_2 = \{0\}$.
Let $x_1,y_1 \in M_1$, $x_2,y_2 \in M_2$ with $T(x_1,x_2) = T(y_1,y_2)$, ie $x_1 + x_2 = y_1 + y_2$.
Then, $x_1 - y_1 = y_2 - x_2 \in M_1\cap M_2 = \{0\}$. Therefore, $x_1= y_1$ and $x_2 = y_2$, i.e. $T$ is injective.
