No surjective bounded linear map from $\ell^2(\mathbf{N})$ to $\ell^1(\mathbf{N})$ The second part of an exercise asks that I show:

There does not exist a surjective bounded linear map from $\ell^2(\mathbf{N})$ to $\ell^1(\mathbf{N})$.  

The first part was to show that these two spaces are separable in their respective $L^p$ norms.  
I've really hit a wall on this one.  I know, in fact, that there does exist a surjective bounded linear map from $\ell^1(\mathbf{N})$ to $\ell^2(\mathbf{N})$.  So my intuition says that if there were also one going the other way, then the two spaces would have to be isomorphic, which is false .  But I don't know of any such theorem that I could apply to "the category of Banach spaces".  
My only other idea is to argue by contradiction using the open mapping theorem, but nothing is coming to me...  My guess is that separability is also going to be important, given the first part of the exercise.
-Thanks.
 A: Hint: Both of these spaces are separable. Note that $\ell^2$ is isomorphic to its dual space, but the dual space of $\ell^1$ fails to be separable.
Note then that a surjection from $\ell^2$ to $\ell^1$ would induce an injection from $(\ell^1)'$ to $(\ell^2)'$.
A: I am going to use the answer suggested by Jochen above, given the unresolved issue of whether the map suggested by Omnomnomnom is open (see the comments).  If this becomes resolved, then I will accept Omnomnomnom's answer over this one.  If you wish to resolve this issue, please do so here:
https://math.stackexchange.com/questions/1317152/is-this-map-of-banach-spaces-open 
Answer:  Suppose $\phi \colon \ell^2 \twoheadrightarrow \ell^1$ is a surjective bounded linear map.  Then $\ell^2 / M \cong \ell^1$ as Banach spaces, where $M = \ker \phi$.  But the quotient of a reflexive normed linear space by a closed subspace is reflexive; e.g. see:
https://books.google.com/books?id=AwHrBwAAQBAJ&pg=PA105&lpg=PA105&dq=quotient+space+is+reflexive&source=bl&ots=0e8I-fhnTm&sig=KAKQpcT3T4ib00XB5SNDkSt0Yog&hl=en&sa=X&ei=iaZ1VeD0HJWvyASe6YGYAg&ved=0CC4Q6AEwAg#v=onepage&q=quotient%20space%20is%20reflexive&f=false
Since $\ell^2$ is reflexive and $\ell^1$ is not, we have a contradiction.  
