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I am running a qualifying exam prep course. A question I posed to my students was:

Suppose that $M$ and $N$ are compact, oriented manifolds and $f:M\longrightarrow N$ is of degree $1$. Show that $f_*:\pi_1(M)\longrightarrow \pi_1(N)$ is a surjection.

A student came up with the following:

There is some covering space, $p:E\longrightarrow N$, of $N$ that corresponds to the subgroup $f_*(\pi_1(M))\leq \pi_1(N)$. The lifting property gives us $\tilde{f}:M\longrightarrow E$ so that $p\circ\tilde{f}=f$. He then argued that $p$ has degree the number of sheets of the covering.

Their definition of degree is to take the preimage of a point $y\in N$ and determine for each $x\in f^{-1}(y)$ whether or not orientation is reversed. This leads to a problem in my student's solution: what if there are an infinite number of sheets in the covering? Then the degree is not really defined. If we had the homology definition or integral definition of degree, this is easily fixed. Is there an argument around this? If not, is there another solution using their definition?

J126
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  • First, I don't see why $f_*(\pi_1(M))$ is necessarily a normal subgroup of $\pi_1(N)$. Second, unless you can argue directly that this subgroup has finite index, I don't see how to avoid a homological argument, looking at $H_2(M)\to H_2(E)\to H_2(N)$. After all, if $f$ had degree $0$, there would be no contradiction [e.g., $S^1 \to \mathbb R \to S^1$]. – Ted Shifrin Jun 25 '13 at 14:07
  • @TedShifrin I never said it had to be a normal subgroup. I also don't see how to avoid a homological argument, hence the question. – J126 Jun 25 '13 at 14:13
  • Oh, sorry, I got messed up on algebra notation I don't usually use :) Are the students not expected to know homology for their quals? :) – Ted Shifrin Jun 25 '13 at 14:19
  • @TedShifrin They are expected to know homology. But, they only learned the degree concept in the oriented manifold setting. They wouldn't have the homology version available to them on the qual. So, I'd rather not use it. – J126 Jun 25 '13 at 19:22
  • Duh. Don't know what I was thinking. Dumb comment deleted. – Jack Lee Jun 26 '13 at 17:52

2 Answers2

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Theorem. Let $M,N$ be closed,connected,oriented,smooth manifolds of the same dimension.Let $f:M\to N$ be a smooth map. Let $\:p:E\to N$ be the covering projection corresponding to the image of $\pi_1(M)$ under the homomorphism induced on fundamental groups by $f$, with $i$ sheets. Let $\:\tilde f:M\to E$ be the lifting of $f$; so $\:f=p\circ\tilde f$.
a) If $i$ is finite, then $\deg f$ is divisible by $i$.
b) If $i$ is infinite, then $\deg f=0$.

Proof without (co)homological tools. a) follows by the theorem on degree of composite maps.
b) Let $y\in N$ be a regular value of $f$. The set $\:Z=\tilde f(M)\cap p^{-1}(y)$ is finite as compact and discrete. Every point $z\in Z$ is a regular value of $\tilde f$ and $p$ is a local diffeomorphism concordant with orientations of $E$ and $N$. The manifold $E$ is not compact, so $\:0=\deg \tilde f =\deg (\tilde f ;z)$ for every $z\in Z$. The set $\:f^{-1}(y)$ is the disjoint union of $\:\tilde f^{-1}(z)$ for $z\in Z$, so $\:\deg f=0$ as the finite sum of $\deg(\tilde f;z)=0$ for $z\in Z$.

Community
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Cezary Bowszyc
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Sorry to revive a seemingly dead question, but here goes. Like the OP said, there is $p:E\rightarrow N$, of $N$ that corresponds to the subgroup $H=f_{*}(\pi_{1}(N)$. The lifting property gives us $\phi:M\rightarrow E$ so that $p\circ \phi=f$.

Since the degree is multiplicative and always an integer, then $deg(p)deg(\phi)=deg(f)=1$ hence the degrees of the two maps must both be $1$ (or $-1$, but that makes no difference). Now, the degree of a covering map is the number of sheets, so $p$ is one sheeted.

Then, the last piece is the theorem that for a covering map, the number of sheets is equal to the index of the subgroup of the covering (theorem 40.G in link below), hence $H$ has index $1$, hence is the whole group.
http://www.pdmi.ras.ru/~olegviro/topoman/eng-book-nopfs.pdf

TheManWhoNeverSleeps
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  • How do you know that $H$ is a finite index subgroup?? – Anubhav Mukherjee Jan 07 '16 at 13:25
  • @Anubhav.K

    Well, I suppose a priori I don't know. However, the manifolds in question (which I assume to be smooth) are oriented. Now, $p$ is a smooth covering map, and $E$ is also oriented. Since $p$ is a local diffeomorphism the Jacobian is nowhere vanishing. One definition of degree is $k=\Sigma sgn(x)$ where $x\in p^{-1}(y)$ where y is some regular value, and $sgn$ is the sign of the Jacobian (which has to be the same everywhere). So, there is only one preimage. Now, we use the theorem I linked in my answer, so the index is 1.

     – TheManWhoNeverSleeps Jan 12 '16 at 08:07