# Are all the smooth fibers in a fibration always homeomorphic?

Let $f:X \rightarrow Y$ be a fibration from a complex manifold $X$ to another connected complex manifold $Y$ such that all the fibers are compact, reduced, connected and smooth. Is it possible that some fiber is not homeomorphic to a generic fiber?

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Do you mean "fibration" in the sense of Hurwicz? It seems very strange to me to use that notion of fibration in the context of complex manifolds... How could you possibly know that a map between complex manifolds is a fibration without actually knowing much more? I guess the question makes sense, but I'm having trouble imagining what context might lead you to that question. – André Henriques Jun 19 '12 at 18:48
Dear Nam--Hoon, If all the fibres are smooth and compact, then under reasonable hypotheses (e.g. we are looking at a map of smooth projective algebraic varieties) then the map itself will be smooth and proper, and so the map will be a proper submersion on underlying smooth manifolds. A proper submersion is always a fibre bundle (Ehresmann) and so the different fibres will be mutually homemorphic. Regards, – Emerton Jun 20 '12 at 14:07
Dear Emerton, Well your answer is what I think but I think that I made a counterexampe. Let $\phi: \widetilde S \rightarrow S$ be universal covering over a Enriques surface $S$. Then $\widetilde$ is a K3 surface. Let $\rho$ be the covering involution over $\widetilde S$ and $\psi:\mathbb P$ be the involtuion, defined by $\psi[x,y] = [-x, y]$. Consider the quotient $X=S \times \mathbb P / \langle \rho, \psi \rangle$ and the projection $\phi:X \rightarrow \mathbb P / \langle \psi \rangle = \mathbb P$. Then two fiber of $phi$ is non-reduced but if one performs a suitable base change, – Nam-Hoon Jun 20 '12 at 16:16
every fiber is reduced and smooth. Then two of the fibers are isomorphic to $S$ but other fibers are isomorphic to $\widetile S$. Note that $S$ and $\widetilde S$ are not homeomorphic. Am I wrong? – Nam-Hoon Jun 20 '12 at 16:18
Dear Nam-Hoon, There are some typos in your comment, but just to check that I understand: do you mean $\widetilde{S} \times \mathbb P^1$ modulo the involution acting on the first factor as $\rho$ and the second factor as $\psi$? Can you tell me what the "suitable base-change" is? Regards, Matthew – Emerton Jun 25 '12 at 2:53