1st Isomorphism Theorem For Banach Spaces: Understanding this proof.

Question

I'm reading Joseph Muscat's "Functional Analysis", and I'm having some trouble understanding the proof of proposition 11.3.

First, let's enumerate the statements in Proposition 11.3, with (i) for the kernel one, (ii) for the statement on closure, (iii) for the last one. At the end of this question, you'll find the propositions and corollary used in the proof.

I'm not getting what he states in the last 2 paragraphs, starting with

«So, J is an isomorphism(...).  [Paragraph 1]

By the Corollary (...). [Paragraph 2]»

In [Paragraph 1], is he trying to prove that (iii)<=> (ii) ? And the inequality referred to is $|x+\text{ker}T|\leq c |Tx|$, right?

In [Paragraph 2], in the first sentence, to use the corollary we must be sure that $X/\text{ker}T$ and $imT$ are banach spaces. However, how do we know that $X/\text{ker}T$ is Banach? We know that $X$ is Banach, but how can we know that $\text{ker}T$ is also Banach? Also, in the same paragraph, 2nd sentence, is he trying to prove (i)=>(ii)? I'm not sure I understand the English.

Below you'll find the propositions and corollary used in the proof, on whose use I have doubts.

Proposition 8.12 is

The Corollary to the Open Map Theorem states «Every bijective operator(linear and continuous map) between Banach spaces is an isomorphism»

Answer 1

A more illuminating version of the statement of Proposition 11.3 is:

Proposition: We have that $(i) \iff (ii)$ and $(i) \iff (iii)$ where

$(i):$ $X / \ker T \cong \operatorname{Im}T$ via the map $J$ (given in the proof).

$(ii):$ $\operatorname{Im} T$ is closed in $Y$.

$(iii):$ $\exists c>0, \forall x \in X \|x + \ker T\| \leq c \|Tx\|$.

With this statement fixed, the proof up to the point "$J$ is an isomorphism precisely when $J^{-1}$ is continuous." is a proof that $(i)$ is equivalent to $J^{-1}$ being continuous. The rest of the proof then shows that conditions $(ii)$ and $(iii)$ are both equivalent to continuity of $J^{-1}$.

In the case of condition $(iii)$, this is immediate by proposition 8.12 and the assumed inequality. Indeed, the assumed inequality is exactly what we need to apply proposition 8.12 to the map $J$. This establishes that $(i) \iff (iii)$

The second paragraph establishes $(i) \iff (ii)$ (or rather $J^{-1}$ continuous $\iff (ii)$). Since $X$ is complete and $\ker T$ is closed in $X$, we know that $\ker T$ is also complete and so $X / \ker T$ is complete by proposition 8.18. If I understand correctly this clears up your issue with the first part of that paragraph which establishes that $(ii) \implies (i)$.

You are correct that the last sentence shows that $(i) \implies (ii)$. The reasoning is that (by the same comments as in my last paragraph) $X / \ker T$ is complete and hence so is any space isomorphic to it. This means that $(i) \implies \operatorname{Im}T$ is complete and so is closed in $Y$.