Every separable Banach space is isometrically isomorphic to a quotient of $\ell^1$ I am currently trying to prove the statement above. So let $X$ be a Banach space and choose a dense sequence $(x_n)_n$ in the closed unit ball of $X$. Then it is easy to see that 
$$T: \ell^1 \to X, \quad Ta = \sum_{n = 1}^\infty a_n x_n$$
is well-defined and $\lVert T \rVert \leq 1$. To prove surjectivity of the map pick $x \in X$ with $\lVert x \rVert \leq 1$. Then by density of the sequence $(x_n)_n$ I can find some $n_1 \in \mathbb N$ such that $\lVert 2x - x_{n_1} \rVert \leq 1$. For the same reason I can find $n_2 \in \mathbb N$ such that $\lVert 2(2x - x_{n_1}) - x_{n_2}\rVert \leq 1$. So inductively I obtain a sequence $(n_k)_k$ such that 
$$ \bigg \lVert x - \sum_{k = 1}^N \frac{1}{2^k} x_{n_k} \bigg \rVert < \frac{1}{2^N}$$
for all $N \in \mathbb N$. Now set $a := \sum_{k = 1}^\infty \frac{1}{2^k} e_{n_k}$ (where $e_n$ denotes the $n$-th unit vector) and obtain that $a \in \ell_1$ and $\lVert a \rVert_{\ell^1} \leq 1$. Moreover, one easily calculates that $Ta = x$. Hence $T$ maps the closed unit ball of $\ell^1$ onto the closed unit ball of $X$. In particular, $T$ is surjective and by the isomorphism theorem 
$$ S: \ell^1/ \ker T \to X,\quad a + \ker T \mapsto Ta $$
is an isomorphism. Hence, $\ell^1/ \ker T \cong X$ as vector spaces. So it is just left to show that $S$ is an isometry. So let $a \in \ell^1$. Then one has clearly that
$$ \lVert S(a + \ker T) \rVert = \lVert T(a + b) \rVert \leq \lVert a + b \rVert_{\ell^1}$$
for each $b \in \ker T$ and therefore
$$ \lVert S(a + \ker T) \rVert \leq \inf \{\lVert a + b \rVert_{\ell^1} : b \in \ker T\} = \lVert a + \ker T \rVert,$$
i.e. $\lVert S \rVert \leq 1$. This means by the inverse theorem that $\ell^1/ \ker T \cong X$ as Banach spaces. But I have no idea how to show that $\lVert S(a + \ker T) \rVert \geq \lVert a + \ker T \rVert$ for all $a \in \ell^1$. 
I think I have to come up with some fancy representative $\tilde a \in \ell^1$ but I do not see how to get that. What am I missing? Please tell me anyone :-)
 A: I don't think this works if you choose "a" dense sequence $\{x_n\}$ in the unit ball of $X$. But here's an idea. Let $A$ be a countable dense subset in the unit ball of $\ell^1$ and $X_0\subset X$ dense in the unit ball of $X$. The set
$$
X_1=\left\{\frac{ \sum_{k=1}^N a_kx_k }{\left\|\sum_{k=1}^N a_kx_k\right\|}:\ \,N\in\mathbb N; x_1,\ldots,x_N\in X,\ a\in A\right\}
$$
is countable and contained in the unit ball of $X$, so $X_0\cup X_1$ is countable and dense, and contained in the unit ball of $X$. Now form the sequence $\{x_n\}$ by repeating each element of $X_0\cup X_1$ infinitely many times. Still countable and dense! 
Define your $T$ using this sequence. Everything you wrote still works. 
Fix $a\in A$, and $\varepsilon>0$. There exists $N_0\in\mathbb N$ with $\sum_{j>N}|a_j|<\varepsilon$ for all $N>N_0$. By the way we constructed the sequence, there exists $m>N$ with $x_m=\frac{ \sum_{k=1}^N a_kx_k }{\left\|\sum_{k=1}^N a_kx_k\right\|}$. Now let $b\in\ell^1$ be the sequence $$ (-a_1,\ldots,-a_N,0,\ldots,\left\|\sum_{k=1}^N a_kx_k\right\|,0,\ldots),$$ where the norm appears in the $m^{\rm th}$ entry. We have 
$$
Tb=-\sum_{k=1}^Na_kx_k+\left\|\sum_{k=1}^N a_kx_k\right\|\,x_m=0,
$$
so $b\in \ker T$. Then
\begin{align}
\|a+\ker T\|&\leq \|a+b\|=\sum_j|a_j+b_j|=|a_m+\left\|\sum_{k=1}^N a_kx_k\right\|\,|\\[0.3cm]
&\leq|a_m|+\left\|\sum_{k=1}^N a_kx_k\right\|\leq\varepsilon+\left\|\sum_{k=1}^N a_kx_k\right\|.
\end{align}
As we can take $N$ arbitrarily big, 
$$
\|a+\ker T\|\leq\limsup_{N\to\infty}\varepsilon+\left\|\sum_{k=1}^N a_kx_k\right\|=\varepsilon+\|Ta\|.
$$
As $\varepsilon$ was arbitrary, we get $\|a+\ker T\|\leq \|Ta\|$. Finally, as this holds in a dense subset of the unit ball of $\ell^1$ and everything is continuous, the inequality holds for any $a$ in the unit ball of $\ell^1$, and by linearity for all $a\in \ell^1$. 
A: First, as Martin Argerami has pointed out, your proof of surjectivity is not completely right. To be more specific, if $\|x\| = 1$, there is no guarantee that you can find some $ n_1 \in \mathbb{N} $ such that $\|2x - x_{n_1}\|\leq 1$.
The condition $\|x\| \leq 1$ needs to be replaced by $\|x\| < 1$. Once this is done, the proof of surjection goes through.
To prove isometry, first notice that $S$, as defined in your proof, is injective, because $S(a + \ker T) = S(b + \ker T)$ implies $a \sim b$.
For any $x = S(a + \ker T)$, if $x=0$, we are done. Otherwise, we can scale it by defining $$x' = \frac{x}{\|x\|_B} (1-\epsilon).$$
Now we define $a'$ from $x'$ as you defined $a$ from $x$. We have the following.

*

*$\|a'\|_{\ell^1}=1$

*$ S\left(\frac{\|x\|_B}{1-\epsilon} a' + \ker T \right) = T\left(\frac{\|x\|_B}{1-\epsilon}a'\right) = \frac{\|x\|_B}{1-\epsilon} x' = x = S\left(a + \ker T \right)$
This implies
$$\|a+\ker T\|_{\ell^1/\ker T } \leq \Big\|\frac{\|x\|_B}{1-\epsilon}a'\Big\|_{\ell^1} = \frac{\|x\|_B}{1-\epsilon}. $$
Since $\epsilon$ can be arbitrarily small, this proves
$$\|a+\ker T\|_{\ell^1/\ker T } \leq \|S(a+\ker T)\|_B.$$
