Let $X$ and $Y$ be Banach space and $W\subseteq Y$ be a subspace. Let $T_1:X\to Y$ be bounded and linear such that $T_1(X)=W$ and $T_2:Y\to Y$ be bounded.
Does there exist a Banach space $Z$ and a surjective bounded operator $T:Z\to\{y\in Y: T_2y\in W\}$?
My attempt: If $T_2$ is bijective, then I know the answer is yes. I couldn't answer the general case.
I considered example of shift and projection operators on $l^\infty$ and could always find a bounded operator but I couldn't generalize it.
Edit: I think I can even handle the case when range of $T_2$ is closed, but I still can't handle the general case.
Best Answer
Yes.
This can be seen as follows:
(1) There exists a complete norm $\|\cdot\|_W$ on $W$ such that the embedding $W \hookrightarrow Y$ is continuous. To see this, use that $T_1$ induces a linear bijection $\tilde T_1: X/\ker T_1 \to W$, which is continuous when $W$ is endowed with the norm from $Y$. Now transport the norm from $X$ to $W$ via the bijection $\tilde T_1$ and denote this new norm on $W$ by $\|\cdot\|_W$; clearly, $W$ becomes complete with this new norm, and the embedding $W \hookrightarrow Y$ is continuous.
(2) Let's use the notation $V := T_2^{-1}(W)$. Now we introduce a new norm $\|\cdot\|_V$ on $V$ by defining $$ \|v\|_V = \|v\|_Y + \|T_2v\|_W. $$ Again, it is not difficult to check that the norm $\|\cdot\|_V$ on $V$ is complete. Hence, we can choose the desired Banach space $Z$ as $(V, \|\cdot\|_V)$ and the desired operator $$ T: Z = (V, \|\cdot\|_V) \to (V, \|\cdot\|_Y) $$ as the identity operator on $V$.