Understanding why $T(X) = Y$ in a proof.

Question

Here is the statement of the theorem:

>

The proof of this direction $(\Rightarrow)$ started with this statement "If $T(X)$ is closed, then its a Banach space with the norm inherited from $Y.$ Thus we may assume that $T(X) = Y.$ "

**But I do not understand why the previous statement will lead us to assume that $T(X) = Y.$ Could anyone explain this for me please? **

Answer 1

Suppose we prove the theorem under the additional assumption that $T(X) = Y$. In other words, we would have proved:

Theorem 3a. Let $X,Z$ be Banach spaces and let $S \in \mathcal{L}(X,Z)$. Suppose $S(X) = Z$. Then $$\exists M > 0 \quad \forall y \in S(X) \quad \exists x \in X \quad S(x)=y\: \text{ and }\: \|x\|\le M \|y\|. \tag{4.1a}$$

Now to actually prove the forward direction of the original Theorem 3, suppose $X,Y$ are Banach spaces, $T \in \mathcal{L}(X,Y)$, and that $T(X)$ is closed. Set $Z = T(X)$ and note this is a Banach space. Let $S=T$, considered as an operator in $\mathcal{L}(X,Z)$; then $S(X) = T(X) = Z$. So $X, Z, S$ satisfy the assumptions of Theorem 3a. Hence $S$ satisfies (4.1a). Now verify (it is completely trivial) that $T$ therefore satisfies (4.1).

In other words, since Theorem 3a with its additional assumption $S(X) =Z$ could be used to immediately prove the original Theorem 3, there was no loss of generality in making that assumption.