TITLE: $\pi_1$ of 4-manifolds that "look like" disk bundles
QUESTION [5 upvotes]: Let $X$ be a smooth compact oriented 4-manifold with $\partial X=L(p,1)$, $H_2(X;\Bbb Z)=\Bbb Z$, $H_3(X; \Bbb Z)=0$ and the induced map $\pi_1(L(p,1)) \to X$ surjective. What are the possibilities for $\pi_1(X)$? In particular, are there examples where $\pi_1 \ne 0$?

REPLY [3 votes]: Let's take your $p$ to be prime. Then $X$ has to be simply connected, even without all of the hypotheses. Here is the argument. 
From the map on $\pi_1(L) \to Z_p$ you get a map $L \to BZ_p$. This map is clearly $0$ in $H_3$, since it factors through the inclusion of $L$ into the $4$-manifold $X$. On the other hand, it's well-known that $L$ generates $H_3(BZ_p) \cong Z_p$. (A quick explanation: you can build $BZ_p$ by attaching a 4-cell by a map of degree $p$ (the universal covering $S^3 \to L$) and then higher cells.)
I think a similar proof works if $p=mn$ with $m$ and $n$ relatively prime, and $\pi_1(X) = Z_n$. You have to know a little more about the map $H_3(BZ_p) \to H_3(BZ_n)$ induced by a surjection. I'm pretty sure it's given by multiplication by $m^2$.