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I would really appreciate if anyone could provide me with an example of a locally trivial, but globally nontrivial, fiber bundle $Y\hookrightarrow Z \rightarrow X$, where $X$, $Y$, and $Z$ are all compact spin manifolds with even dimension (preferably divisible by $4$) satisfying $p_1(Z)=0$, but $p_1(X)\neq 0$.

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This question has been answered on mathoverflow… – clh Jul 29 '11 at 2:47

Sorry to be so late to the party, but I wanted to mention a general construction for these kinds of problems.

Given a compact Lie group $G$ and a closed subgroup $H\subseteq G$, the action of $H$ on $G$ by (say) right multiplication is free so gives rise to a principal $H$ bundle $H\rightarrow G\rightarrow G/H$.

Since $H$ and $G$ are Lie groups, they are parallelizable, so $H$ and $G$ are spin, have $p_1 = 0$, etc.

On the other hand, there are fairly straight forward methods of computing the topology (homotopy groups, cohomology ring, Stieffel-Whitney and Pontryagin classes) in terms of those of $H$ and $G$ and the embedding $H\rightarrow G$ (which are, at least when $H$ and $G$ are simply connected, quite well understood).

More generally, one can pick 2 homomorphisms $H\rightarrow G$ (equivalently, one homomorphism $H\rightarrow G\times G$ and use this to define and action on $G$ via $(h_1,h_2)*g = h_1gh_2^{-1}$. This generalizes the previous construction (which occurs when the first homorphism is trivial). These examples, when free, still have topology which is relatively computatable, but give an even larger class of examples to work with.

Then, for example, a simple example of what you're looking for is $U(2)\subseteq SU(3)$. Here, $U(2)$ denotes the set of $2\times 2$ complex unitary matrices and $SU(3)$ denotes the set of $3\times 3$ complex unitary matrices with determinant $1$. The inclusion sends $A$ to $\text{diag}(A,\overline{\det(A)})$.

It is fairly well known that in this case $SU(3)/U(2) = \mathbb{C}P^2$, which has nontrivial $p_1$.

So the bundle $U(2)\rightarrow SU(3)\rightarrow \mathbb{C}P^2$ is an example of what you're looking for. Further, it's nontrival as $SU(3)$ is simply connected but $\pi_1(U(2)\times \mathbb{C}P^2) = \mathbb{Z}$.

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