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Suppose that $M$ and $B$ are two smooth manifolds and $\Pi:M\rightarrow B$ a submersion (and onto). Fixed $x\in M$ and $b\in B$, is the induced homomorphism $\Pi_{\#}:\pi_{1}(M,x)\rightarrow \pi_{1}(B,b)$ also onto? I think so if we assume that the fibres are connected, but I am not sure.

This is my argument. If $\gamma:[0,1]\rightarrow B$ is a parametrization of an element of $\pi_1(B,b)$, using that $\Pi$ is a submersion, it is clear that locally we can lift $\gamma$ to $M$, obtaining a finite number of curves $\alpha_i:J_i\rightarrow M$ such that $\Pi(\alpha_i(t))=\gamma(t)$ for all $t\in J_i$, where $J_i$ are closed intervals with $\cup_i J_i=[0,1]$ and $J_i\cap J_{i+1}=\{one point \}$. The problem is that $\alpha_i$ does not need to glue with $\alpha_{i+1}$, but this can be solved easily taking into account that the end point of $\alpha_i$ and the starting point of $\alpha_{i+1}$ are in the same fibre.

Indeed. Since we assume that the fibres are connected, we can take an auxiliary curve $\beta_i$ in the fibre joining the end point of $\alpha_i$ and the starting point of $\alpha_{i+1}$. Composing, we obtain a loop $\sigma$ in $M$ such that $\Pi_{\#}([\sigma])=[\gamma]$.

Thanks in advance.

benji
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    This is not true! Try $\pi:\Bbb{R}\to\Bbb{S}^1$ defined by $\pi(t)=e^{it}$, this is a onto submersion, but the induced homomorphism is not onto. – DiegoMath Jul 23 '14 at 22:53
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    @DiegoMath: The OP stipulated that the fibers are connected. – Jack Lee Jul 23 '14 at 23:38
  • I think you need a little more justification for the existence of the $\alpha_i$. You could use for example the regular value theorem (which maybe you implicitly are). – Eric O. Korman Jul 24 '14 at 00:15
  • I haven't checked your proof but I have provided an answer confirming the result under the topological supplementary hypothesis of properness. – Georges Elencwajg Jul 24 '14 at 09:37

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If $\Pi:M\to B$ is a fiber bundle, we associate to it a long exact sequence of homotopy groups, a fragment of which is $$ \cdots \to \pi_1(M,x)\to \pi_1(B,b)\to \pi_0(F,x)\to \cdots $$ where $F=\Pi^{-1}(b)$ is the fiber over $b$.
Hence if $F$ is connected, then $\pi_0(F,x)=0 $ and we indeed have surjectivity of $ \pi_1(M,x)\to \pi_1(B,b) $.

But is $\Pi$ a fibration when $\Pi$ is a submersion?

The answer is yes if we assume thar $\Pi$ is a proper map: this is Ehresmann's theorem.
Summing up, we have in your situation the desired surjectivity on fundamental groups under the supplementary topological hypothesis of properness.

Georges Elencwajg
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    Thanks you all. Georges Elencwajg, I have already known the Ehresmann's theorem, but I would like to know if the surjectivity is assured for any submersion. I have not found this result, but I think it is quite direct. – benji Jul 24 '14 at 12:28