Problem 5.3.35 from Folland: Let $\mathscr{X}$ and $\mathscr{Y}$ be Banach spaces, $T\in L(\mathscr{X},\mathscr{Y})$, $\mathscr{N} =\{x: Tx = 0\}$, and $\mathscr{M} = range(T)$. Then $\mathscr{X}/\mathscr{N}(T)$ is isomorphic to $\mathscr{M}$ iff $\mathscr{M}$ is closed.
Normally I would of attempted to prove this but I am pretty lost on where to start, I think my first step should be to show that $T$ is bijective then that would imply $T$ is an isomorphism but I am not sure. Any suggestions is greatly appreciated. I will edit this post if I come up with anything.
I believe it's an application of the open mapping theorem. I'm a bit new to Functional analysis but I think this is the reasoning:
Let $X$ and $Y$ be Banach spaces and $T\in L(X,Y)$. Then $$T(X) \text{ is closed } \iff X/\ker T \cong T(X).$$
(Range of $T$ is denoted as $T(X)$ and kernel of $T$ is the null space of $T$)
Proof: Suppose $T(X)$ is closed. Define $\phi: X/\ker T \to Y$ as $\phi(x+\ker T)=Tx$. Now $\phi$ is a (well defined) bounded linear bijective mapping from $X/\ker T \to T(X)$ (why?) and since $Y$ is complete, and $T(X)\subset Y$ a closed subspace, we know that $T(X)$ is complete. Thus by the open mapping theorem, $\phi$ is an isomorphism of $X/\ker T \to T(X)$.
Conversely, if $X/\ker T \cong T(X)$ then $T(X)$ is complete since it is isomorphic to $X/\ker T$, which is also a Banach space (why?), hence $T(X)$ is closed.