A quotient space is isomorphic if we have closedness 
Let $\mathcal{X}$ and $\mathcal{Y}$ be Banach spaces, $T\in L(\mathcal{X},\mathcal{Y})$, $\mathcal{N} =\{x: Tx = 0\}$, and $M = range(T)$. Then $\mathcal{X}/\mathcal{N}(T)$ is isomorphic to $\mathcal{M}$ iff $\mathcal{M}$ is closed.

I know that by definition $T(x)$ is closed, it seems pretty obvious that $M$ is then closed unless I am missing something but I don't grasp the idea of how the quotient space $\mathcal{X}/\mathcal{N}(T)$ is isomorphic to $M$ depending on $M$ being closed. 
Any ideas or suggestions is greatly appreciated I am really confused with this one. 
 A: One side is trivial: if the two spaces are isomorphic, then $\mathcal{M}$ has to be closed since $\mathcal{X}/\mathcal{N}(T)$ is closed.
Fundamental algebra shows $\mathcal{X}/\mathcal{N}(T)\cong\mathcal{M}$ as linear spaces. More precisely, $T$ induces a canonical bijective linear operator $\bar{T}:\mathcal{X}/\mathcal{N}(T)\rightarrow\mathcal{M}$ by $\bar{T}:x+\mathcal{N}(T)\mapsto T(x)$. It's plain that $\bar{T}$ is continuous by checking quotient topology. But there is still one more step to go to make $\bar{T}$ an isomorphism.
What role does $\mathcal{M}$ being closed play? One may recall the open mapping theorem. If $\mathcal{M}$ is complete, thus a Banach space, then $\bar{T}$ is an open map, hence an isomorphism. But any closed subspace of a complete space is complete. Now we're done.
A: 
Let $\mathcal{X}$ and $\mathcal{Y}$ be Banach spaces, $T\in L(\mathcal{X},\mathcal{Y})$, $\mathcal{N}(T) =\{x: Tx = 0\}$, and $\mathcal{M} = \textrm{range}(T)$. Then $\mathcal{X}/\mathcal{N}(T)$ is isomorphic to $\mathcal{M}$ iff $\mathcal{M}$ is closed.

Proof :
($\Rightarrow$) Suppose that $\mathcal{X}/\mathcal{N}(T)$ is isomorphic to $\mathcal{M}$. By Exercise 15 a.) (in Folland's book), we know that $\mathcal{N}(T)$ is a closed subspace of  $\mathcal{X}$. By Exercise 12 d.), we know that $\mathcal{X}/\mathcal{N}(T)$ is a complete normed vector space, that is, it is a Banach space. 
Let $S : \mathcal{X}/\mathcal{N}(T) \to \mathcal{M}$ be an isomorphism. Let us prove that $\mathcal{M}$ is closed in $\mathcal{Y}$. 
Let $\{y_j\}_{j\in \mathbb{N}}$ in $\mathcal{M}$ such that $y_j \to y \in \mathcal{Y}$.  Then $\{y_j\}_{j\in \mathbb{N}}$ is a Cauchy sequence. Since $S^{-1}$ is continuous linear mapping, it is uniformly continuous and so $\{S^{-1}(y_j)\}_{j\in \mathbb{N}}$ is a Cauchy sequence in $\mathcal{X}/\mathcal{N}(T)$. Since $\mathcal{X}/\mathcal{N}(T)$ is complete, there is $x \in \mathcal{X}/\mathcal{N}(T)$ such that $S^{-1}(y_j) \to x$. Since $S$ is continuous, we have
$$ y_j = S(S^{-1}(y_j)) \to S(x)$$ 
So we have that $y=S(x)$, which means $y \in \mathcal{M}$. So we proved that $\mathcal{M}$ is closed in $\mathcal{Y}$. 
($\Leftarrow$) Suppose $\mathcal{M}$ is closed. Then we have that $\mathcal{M}$ (with the norm of $\mathcal{Y}$) is a complete normed vector space, that is, $\mathcal{M}$ is a Banach space. 
Let $S : \mathcal{X}/\mathcal{N}(T) \to \mathcal{M}$, defined by $S(x+\mathcal{N}(T))=T(x)$.  By Exercise 15 b.) (in Folland's book), we know that $S \in L(\mathcal{X}/\mathcal{N}(T), \mathcal{Y})$ and so $S \in L(\mathcal{X}/\mathcal{N}(T), \mathcal{M})$. 
It is easy to prove that $S$ is bijective. In fact, if $S(x+\mathcal{N}(T))=0$, then $T(x)=S(x+\mathcal{N}(T))=0$, then $x\in \mathcal{N}$, which means $x+\mathcal{N}(T)=0+\mathcal{N}(T)$. So $S$ is injective. Since $\textrm{range}(S)=\textrm{range}(T)=\mathcal{M}$, $S$ is surjective.
Now, apply the Inverse Mapping Theorem (corollary 5.11 in Folland's book) and we can conclude that $S$ is an isomorphism. 
(Or, if you prefer, you can apply the Open Mapping Theorem, theorem 5.10 in Folland's book, to conclude that  $S$ is an isomorphism.) 
Remark: Note that in ($\Rightarrow$) part, the fact that the isomorphism $S$ is linear has a critical role, so we can, from continuity of $S^{-1}$, ensure $S^{-1}$ is uniformly continuous.  
Compare to the following exemple: $f: \mathbb{R} \to (-\frac{\pi}{2},\frac{\pi}{2})$, defined by $f(x) =\arctan (x)$. Note that $f$ is a continuous bijective function whose inverse is continuous, but $(-\frac{\pi}{2},\frac{\pi}{2})$ is not closed in $\mathbb{R}$.
