The inner product of the Cartesian Product space

One can check directly that setting $$ \langle (a_1, b_1), (a_2, b_2) \rangle _{A \times B}= \langle a_1 , a_2 \rangle_A + \langle b_1 ,b_2 \rangle_B $$ defines an inner product on $A\times B$. Linearity and switching properties are easy, and also that $$ \langle (a,b),(a,b)\rangle_{A\times B}\ge0 $$ Now, suppose that $\langle (a,b),(a,b)\rangle_{A\times B}=0$. Then $\langle a,a\rangle_A+\langle b,b\rangle_B=0$, from which $a=0$ and $b=0$ follows.

The projection maps $p_A\colon A\times B\to A$ and $p_B\colon A\times B\to B$ are bounded (that is, continuous) and it's readily shown that this is a product in the sense that if we are given a Hilbert space $C$ and bounded linear maps $f_A\colon A\to C$, $f_B\colon B\to C$, there is a unique bounded linear map $g\colon A\times B\to C$ such that $f_A=p_A\circ g$ and $f_B=p_B\circ g$: just define $$ g(a,b)=f_A(a)+f_B(b) $$ and check boundedness.