Surjective function on product space I know that, if $U$ and $V$ are closed subspaces of a Hilbert $(X,\langle\cdot,\cdot\rangle)$, then these statements are equivalents:
$$i)\ U^{\perp}\subseteq U+V\quad\quad\quad ii)\ X = U+V\quad\quad\quad iii)\ U^{\perp}+V^{\perp} \subseteq U+V.$$
Now, I want to prove this:

Let $X,Y_1, Y_2$ be Hilbert spaces, and $Y = Y_1\times Y_2$. Let $\mathcal{B}\in\mathcal{L}(X,Y)$ such that $\mathcal{B}(x) = \left(\mathcal{B}_1(x), \mathcal{B}_2(x)\right)^T$, $\forall x\in X$, where $\mathcal{B}_i\in\mathcal{L}(X,Y_i)$, $\forall i\in\{1,2\}$. Show that $\mathcal{B}$ is onto if and only if
$$i)\ \mathcal{B}_1 \mbox{ and } \mathcal{B}_2 \mbox{ are onto}.\quad\quad\quad ii)\ X = \mbox{Ker}(\mathcal{B}_1) + \mbox{Ker}(\mathcal{B}_2).$$

Please can sombody help me with this problem?
Thanks in advance
 A: *

*First, suppose that $\mathcal{B}$ is onto. Let $\pi_1: Y_1 \times Y_2 \to Y_1$ and $\pi_2: Y_1 \times Y_2 \to Y_2$ be the orthogonal projections onto the factors $Y_1$ and $Y_2$ of $Y_1 \times Y_2$, which are surjective. Then $\mathcal{B}_1 = \pi_1 \circ \mathcal{B}$ and $\mathcal{B}_2 = \pi_2 \circ \mathcal{B}$ are compositions of surjections, and so... On the other hand, let $x \in X$, and by surjectivity of $\mathcal{B}$, let $x^\prime \in X$ be such that $\mathcal{B}(x^\prime) = (0,\mathcal{B}_2(x))$. Then $\mathcal{B}_1(x^\prime) = 0$ and $\mathcal{B}_2(x-x^\prime) = \mathcal{B}_2(x)-\mathcal{B}_2(x^\prime) = 0$, so that $x = x^\prime + (x-x^\prime) \in \ker \mathcal{B}_1 + \ker \mathcal{B}_2$.

*Now, suppose that $\mathcal{B}_1$ and $\mathcal{B}_2$ are onto and that $X = \ker\mathcal{B}_1 + \ker\mathcal{B}_2$. Let $(y_1,y_2) \in Y_1 \times Y_2$, so that $y_1 = \mathcal{B}_1 x_1$ and $y_2 = \mathcal{B}_2 x_2$ for some $x_1$ and $x_2 \in X$. Since $X = \ker \mathcal{B}_1 + \ker \mathcal{B}_2$, write,
$$
 x_1 = x_1^\prime + x_1^0, \quad x_1^\prime \in \ker \mathcal{B}_2, \; x_1^0 \in \ker \mathcal{B}_1,\\
 x_2 = x_2^\prime + x_2^0, \quad x_2^\prime \in \ker \mathcal{B}_1, \; x_2^0 \in \ker \mathcal{B}_2,\\
$$
and observe that 
$$
 y_k = \mathcal{B}_k(x_k) = \mathcal{B}_k(x_k^\prime+x_k^0) = \mathcal{B}_k(x_k^\prime) + \mathcal{B}_k(x_k^0) = \mathcal{B}_k(x_k^\prime), \quad k=1,2.
$$
What, then, is $\mathcal{B}(x_1^\prime+x_2^\prime)$?

