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The situation I find myself in is as follows: I have a CW complex $X$ which is covered by two subcomplexes $A$ and $B$ and I know that $A$, $B$ and $A \cap B$ are connected and aspherical. The term aspherical means that all higher homotopy groups $\pi_i(-)$ for $i\geq 2$ vanish. Recall that the Seifert van Kampen theorem implies that the fundamental group of $X$ is the amalgamation $A\ast_{A\cap B} B$.

Now I had thought that the inclusions $\pi_1(A\cap B \rightarrow A)$ and $\pi_1(A\cap B\rightarrow B)$ were both injective. In this situation there is a Theorem which states that $X$ is itself aspherical, and so its homotopy type is described by its fundamental group $A\ast_{A\cap B} B$. However I have now learnt that my maps are not injective, however I still have strong reason to believe that my space $X$ is aspherical.

I can prove that $\pi_1(A\cap B\rightarrow A\times B)$ is injective, so I want the following to hold:

Theorem(?)

Suppose that $X$, $A$, $B$ and $A\cap B$ are as above and that $\pi_1(A\cap B\rightarrow A\times B)$ is injective. Then $X$ is aspherical.

I can sketch a proof: the main step is to show that any $\phi\in\pi_2(X;A\cap B)$ can be decomposed into a product of $\phi_A\in\pi_2(A;A\cap B)$ and $\phi_B\in\pi_2(B;A\cap B)$. Then in the long exact sequence of homotopy groups associated to $(X;A\cap B)$ the element $\phi$ comes from $\pi_2(X)$ only if the images of $\phi_A$ and $\phi_B$ in $\pi_1(A\cap B)$ are inverse. But these images must be in the kernels of the respective maps into $\pi_1(A)$ and $\pi_1(B)$, hence by the injectivity assumption are trivial. But now the long exact sequences for $(A;A\cap B)$ and $(B;A\cap B)$ imply that $\phi_A$ and $\phi_B$ are null homotopic and hence so is $\phi$.

But I have not been able to find the result in the literature.

Questions

  1. Is this theorem true?
  2. Is there a reference for this result?
  3. Are there any well known examples of this situation? Clearly any amalgamation of groups can be converted to one where the maps are injective, however the groups obtained in this way may not be themselves as nice as those with which we started. In my case the fundamental groups go from being finitely presented to only being finitely generated.
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I've realised that I've used the term amalgamation, where perhaps I should have used amalgamated free product. Or just pushout. –  James Griffin May 23 '11 at 17:18
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1 Answer

up vote 3 down vote accepted

Counterexample: The union of $A=D^2\times S^1$ and $B=S^1\times D^2$ is homeomorphic to $S^3$.

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Thank you, I really should have checked all of the long exact sequence, I might have spotted that one. Although it also fails for the pi_2 part as well in an even simpler way; take B to be the disc D_2 with S^1 as boundary. Then take A to be the torus with a radial circle. Whoops. The pictures for how my 'proof' fails are however very pretty. Back to the drawing board... –  James Griffin May 23 '11 at 18:28
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