Exercise 5.36 in Folland's Real Analysis reads as follows:
Let $X$ be a separable Banach space and let $\mu$ be the counting measure on $\mathbb N$. Suppose that $\{x_n\}$ is a countbale dense subset of the unit ball of $X$ and define $T:L^1(\mu) \to X$ as $$ Tf = \sum_1^\infty f(n) x_n. $$ a. Show that $T$ is bounded. b. Show that $T$ is surjective. c. Show that $X$ is isomorphic to a quotient space of $L^1(\mu)$.
My question regards item c. Folland gives a hint of looking at exercise 5.35, which reads as
Let $X$ and $Y$ be Banach spaces, $T \in L(X, Y)$, $N(T) = \{x : Tx = 0\}$ and $M =$ range$(T)$. Show that $X/N(T)$ is isomorphic to $M$ iff $M$ is closed.
Looking at exercise 35, the first thing that comes to mind in to take $N(T)$ for $T$ defined in exercise 36. However, this seems too simple. What exactly does he want, and how do we proceed? Do we even know if $N(T)$ is not trivial?
Thanks in advance and kind regards.
Part(c) follows follows from (a) and (b) by the hint offered in the OP with $T(L_1(\mu))=X$ in lieu of $M$.
Here is a short proof that this is the case:
Fist by (a) $T$ is continuous and so $N_T=\{f\in L_1(\mu): Tf=0\}$ is a close subset of $L_1(\mu)$. Being that $L_1(\mu)$ is Banach space, we have that $L_1(\mu)/N_T$, with the quotient topology, is a Banach space under the norm $$\|f^*\|:=\inf_{g\in N_T}\|f-g\|_{L_1(\mu)}$$ Let $\pi:L_1(\mu)\rightarrow L_1(\mu)/N_T$ be the quotient map.
By (b) $T$ is surjective, by a well known result, there exists a unique continuous map $$\phi:L_1(\mu)/N_T\rightarrow Y$$ such that $\phi\circ \pi =T$, namely $$\phi(f+N_T)=Tf$$ It is easy to check that $\phi$ is linear, bounded and one to one and onto. The continuity of $\phi^{-1}$ follows from the open map theorem (Here we use the fact that $Y$ is a Banach space).
Here we furnish short proofs that (a) and (b) hold.
(a) $\|Tf\|_X\leq \sum_n|f(n)|\|x_n\|_X\leq \sum_n |f(n)|=\|f\|_1$
(b) is a little tricky. Here is one possible solution.
For each $m\in\mathbb{N}$, define $e_m(n)=\mathbb{1}_{\{m\}}(n)$. Clearly $e_m\in L_1(\mu)$, $\|e_m\|_{L_1(\mu)}=1$, and $Te_n=x_n$.
It is enough to show that $B_X(0;1)\subset T(L_1(\mu))$. Let $y_1=y\in B(0;1)$. There exists $x_{n_1}$ such that $$ \big\|\tfrac{y_1}{\|y_1\|_X} -x_{n_1}\big\|<\frac12 $$ and so, $$ \big\|y_1-\|y_1|_Xx_{n_1}\big\|_X\leq\frac12\|y_1\|_X\leq\frac12 $$ Set $y_2=y_1-\|y\|_Xx_{n_1}$. Then, $\|y_2\|_X\leq\frac12$, and there is $x_{n_2}$ such that $$ \big\|\tfrac{y_2}{\|y_2\|_X} -x_{n_2}\big\|<\frac{1}{2^2}, $$ which in turn implies that $$\big\|y_1-\|y_1\|_Xx_{n_1}-\|y_2\|_Xx_{n_2}\big\|_X=\big\|y_2-\|y_2\|_Xx_{n_2}\big\|_X\leq\frac{1}{2^2}\frac{1}{2}=\frac{1}{2^{2+1}} $$ proceeding by induction, we obtain sequences $\{y_k:k\in\mathbb{N}\}$ and $\{x_{n_k}:k\in\mathbb{N}\}$ such that $$ y_k=y_{k-1}-\|y_{k-1}\|_Xx_{n_{k-1}}=y_1-\Big(\|y_{n_{k-1}}\|_Xx_{n_{k-1}} +\ldots+ \|y_1\|_Xx_{n_1}\big), $$ $$ \|y_k\|_X\leq \frac{1}{2^{(k-1)+\ldots+1}}, $$ and $$ \big\|\tfrac{y_{n_k}}{\|y_{n_k}\|_X} -x_{n_k}\big\|<\frac{1}{2^k}, $$ which in turn implies that $$ \big\|y_{n_{k+1}}\big\|_X=\big\|y_k-\|y_{n_k}\|x_{n_k}\big\|_X\leq \frac{1}{2^k}\|y_{n_k}\|_X\leq \frac{1}{2^{k+(k-1)+\ldots+1}} $$ It follows that $f=\sum^\infty_{k=1}\|y_k\|_Xe_{n_k}\in L_1(\mu)$ since $$ \sum^\infty_{k=1}\big\|\|y_k\|_Xe_{n_k}\big\|_{L_1(\mu)}=\sum^\infty_{k=1}\|y_k\|_X<\sum^\infty_{k=1}2^{-k}=1 $$ Furthermore, $$ Tf=\sum^\infty_{k=1}\|y_k\|_Xx_{n_k}=x $$ Therefore $\overline{B}_X(0;1)\subset T\big(B_{L_1(\mu)}(0;1)\big)$.