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 $\pi_2(X)\ne 1$?
Let $X$ be a CW-complex with
Is it always true that $\pi_2(X)\ne 1$?
There are classic examples, coming from small cancellation theory. See the section of the Wikipedia article on asphericity.
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.
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:
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? ;-)