I have an exercise on group isomorphisms but I do not quite understand the second part of their proof.
Assume that $B$ is a symmetric bilinear form, and let $q : V \to \mathbb{R}$ be the map given by $q(u) = \frac{1}{2}B(u, u)$.
Show that the map $\phi : G \to G$, given by $\phi(u, a) = (u, a + q(u))$ for $u \in V$ and $a \in \mathbb{R}$ is a group-isomorphism.
I have showed that it is compatible with composition laws. For bijectivity:
Injectivity I could show by showing $\ker(\phi)=(0,0)$,
so $(u,a+\frac{1}{2}B(u, u))=(0,0) \Rightarrow u = 0$
and $a+\frac{1}{2}B(u, u)=0$
Since $u = 0$ we know that $\frac{1}{2}B(u, u) = 0$ and therefore $a = 0$
For surjectivity however, I do not quite understand the proof.
Since $\phi(u,a-q(u))=(u,a)$, every element of $G$ is in the image of $\phi$.
What are the steps to get this solution? Is there some intermediate step that could help?
By definition of $\phi$, we have $$ \phi(u,a-q(u)) = \big(u,(a-q(u))+q(u)\big) = (u,a). $$ It remains to show that $(u,a-q(u)) \in G$, but I'm not sure how $G$ is defined, so give that a try yourself! What you have then shown is that every element of $G$ is the image of some other element in $G$ under $\phi$.