In order to prove the following statement:
"Let $F: U\to V$ be a linear map, and assume that this map has an inverse mapping $G\colon V \to U$. Then $G$ is a linear map."
In Serge Lang book he chosed two arbitrary elements of $V$, namely $v_1$ and $v_2$ after to prove that $G( v_1 + v_2 )=G(v_1)+G(v_2)$ he called $G(v_1)=u_1$ and $G(v_2)=u_2$ after he wrote:
$$\tag{I}"F(G(v_1)+G(v_2))=F(u_1+u_2)=F(u_1)+F(u_2)= v_1 + v_2 = F(G(v_1 + v_2))" $$
and from here he concluded that
$$G( v_1 + v_2 )=G(v_1)+G(v_2).\tag{II}$$
My point is why he could do this conclude $(II)$ knowing $(I)$, because if the image of two elements are equal not necessarily requires that those two elements are equals, moreover if he implicitely used the necessity to be injective to exist a inverse how can one show that $F$ is injective if $F$ accept a inverse?
Thank You.
Because $F$ has an inverse, $F$ is injective--that is, there do not exist distinct $u_1,u_2\in U$ such that $F(u_1) = F(u_2)$. Otherwise, the inverse map $G$ would not be well-defined, since $G(F(u_1))$ "could" be either $u_1$ or $u_2$. Thus $F(u_1) = F(u_2)$ implies $u_1 = u_2$. This is the step that takes us from $(I)$ to $(II)$.