Let $X = Y \cup Z$ be a connected, path-connected Hausdorff space. Suppose that $Y$, $Z$, and $Y\cap Z$ are all connected, path-connected, and aspherical, and that the homomorphism $\pi_1(Y\cap Z) \to \pi_1(Y)$ induced by inclusion is injective.

Does it follow that $X$ is aspherical? If so, why?

Edit: This statement is intuitively true, in the sense that a similar statement holds for homology. That is, if we know that $H_n(Y) = H_n(Z) = H_n(Y\cap Z) = 0$ for all $n\geq 2$, and the homomorphism $H_1(Y\cap Z) \to H_1(Y)$ is injective, then it follows that $H_n(X)=0$ for all $n \geq 2$. This is a consequence of the Mayer-Vietoris sequence: $$ \cdots \to H_3(Y)\oplus H_3(Z) \to H_3(X) \to H_2(Y\cap Z) \to H_2(Y)\oplus H_2(Z) \to H_2(X) \to H_1(Y\cap Z) \xrightarrow{\varphi} H_1(Y) \oplus H_1(Z) $$ Plugging in $H_n(Y)=H_n(Z) = H_n(Y\cap Z) = 0$ for $n\geq 2$, we get $$ \cdots \to 0 \to 0 \to H_3(X) \to 0 \to 0 \to H_2(X) \to H_1(Y\cap Z) \xrightarrow{\varphi} H_1(Y) \oplus H_1(Z) $$ Clearly $H_n(X) = 0$ for $n\geq 3$. Moreover, since the homomorphism $H_1(Y\cap Z)\to H_1(Y)$ is the first coordinate of $\varphi$, we know that $\varphi$ is injective, and therefore $H_2(X) = 0$ as well.

The problem is that there's no good analog of the Mayer-Vietoris sequence for homotopy groups, and the versions of excision that I'm aware of only work in the case where the two component spaces are $n$-connected. Here the spaces $Y$ and $Z$ may have nontrivial fundamental groups, and it's not clear to me how to resolve the issue using covering spaces.

By the way, I'm willing to make more topological assumptions about the spaces $X$, $Y$, and $Z$. For example, they are certainly semi-locally simply connected and locally compact, and they probably have the homotopy type of CW complexes.

  • $\begingroup$ Have you done any work of your own? $\endgroup$
    – user641
    Sep 30, 2013 at 19:01
  • $\begingroup$ @SteveD This isn't homework, if that's what you're asking. I'm trying to understand a claim in a paper that I'm reading. I looked through chapter 4 of Hatcher's book for a while, and I couldn't find any theorems that immediately applied, though maybe I'm overlooking something obvious. $\endgroup$
    – Magritte
    Sep 30, 2013 at 19:04
  • $\begingroup$ No, I never suggested it was homework. But questions posted here without context or work usually get closed. $\endgroup$
    – user641
    Sep 30, 2013 at 19:05
  • $\begingroup$ @SteveD Thanks for the concern. Based on the way this is treated in the paper, I'm assuming it's some simple argument, but I'm just not seeing it. Hopefully someone here can help! $\endgroup$
    – Magritte
    Sep 30, 2013 at 19:06
  • $\begingroup$ Anyway, you might try first the slightly simpler problem where $Y$, $Z$, and $Y\cap Z$ are contractible. Then consider $X=S^2$ and $Y$ and $Z$ the hemispheres, and see what fails there (in other words, why the injectivity hypothesis is crucial). $\endgroup$
    – user641
    Sep 30, 2013 at 19:07

1 Answer 1


This question has been asked and answered on MathOverflow. I have replicated the accepted answer by Eric Wofsey below.

This is false in general. Let $Y=S^1\times S^1$, and let $Z$ be a disk whose boundary is identified with $S^1\times\ast$. Then $Y\cup Z\simeq S^1\vee S^2$ is not aspherical. Are there any additional conditions in the situation you care about that fail for this example?


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