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.
