The second homotopy group of a simple CW-complex Let $X$ be a CW-complex with 


*

*one 0-cell

*two 1-cells

*three 2-cells

*no cells in dimensions 3 or higher. 


Is it always true that $\pi_2(X)\ne 1$?
 A: I believe that the answer is NO. If you look at 
Gutiérrez, Mauricio A.; Ratcliffe, John G.
On the second homotopy group. 
Quart. J. Math. Oxford Ser. (2) 32 (1981), no. 125, 45–55. 
Corollary 3 states that a "reduced 2-complex $K(X; R)$ is aspherical if and only if each element of $R$ is independent and not a proper power."
Now, "reduced" means that there is (a) only one 0-cell (true in your case), and the one cells represent distinct nontrivial elements of $\pi_1(K^1),$ where $K^1$ is the one-skeleton. Again seems to be true under your assumptions. $R$ are the relations (given by attaching maps of the 2-cells, I imagine), "independent" is too complicated to explain here (look at the paper), but in any case, the "not a proper power" condition is easy to violate.
EDIT Actually, independent is not too hard to explain. The definition is: a relator $r$ is independent if, setting $M$ to be the normal closure of $r,$ and $N$ the normal closure of $R - r,$ $M \cap N = [ M, N].$
As @Benjamin points out, above I am answering the complementary question, so to get the example that the OP wants, we need three independent elements in the free group on two generators which are not proper powers.
A: There are classic examples, coming from small cancellation theory. See
the section of the Wikipedia article on asphericity. 
A: So, the one 0-cell forces the 1-skeleton to be a figure-8. And we attach three 2-cells to this figure-8. These cells can be attached to:


*

*loop 1, with some winding number n.

*loop 2, with some winding number m.

*the 0-cell, and it's degenerate 1-cell. 


In the last case, we get a generator for $\pi_2$ from the resulting sphere; and without any 3-cells, any generator that shows up will produce non-trivial homotopy.
Suppose, thus, that the last case does not occur. Then we would be distributing three 2-cells on 2 loops. Regardless of how we do this, at least two 2-cells attach to the same loop, possibly with different winding numbers. Unless all three 2-cells attach to the same loop, the fundamental group will be trivial. If $\pi_1$ is indeed trivial, then because $H_2(X)=Ab \pi_2(X)$, it follows that $\pi_2(X)$ is indeed non-trivial. If all three 2-cells attach to the same loop, then the space is a wedge of a circle and the CW-complex on 1 0-cell, 1 1-cell and 3 2-cells. Being a wedge, if the homotopy on a factor is non-trivial, the entire homotopy will also be, and for the factor of the three attached 2-cells, the above argument with the abelianization also works out.
... or at least, that's how I would approach it. Would those here who know homotopy theory now please tell me why this cannot possibly work? ;-)


