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This is a question from a past qualifying exam I am stuck on:

For a smooth map $f:M\rightarrow N$ between smooth, compact, oriented $n$-manifolds, the degree of $f$ is the unique integer $k$ such that:

$$\int_M f^*\omega = k\int_N\omega$$

for every smooth $n$-form $\omega$ on $N$. Prove or disprove the following:

i) There exists a degree-1 smooth map $S^2 \times S^3 \rightarrow S^5$

ii) There exists a degree-1 smooth map $S^5 \rightarrow S^2\times S^3$.

There was a similar problem from another exam that I got (that there is no degree-1 map from $S^2 \rightarrow T^2$) but the same trick doesn't seem to work here.

There doesn't exist a degree-1 map from $S^2$ to $T^2$:

Proof: Suppose $f:S^2 \rightarrow T^2$ is a degree-1 map. Then $f$ induces an isomorphism $f^*:H^2(T^2)\rightarrow H^2(S^2)$ (this is actually the definition used in the other problem but it is easy enough to see that the two are equivalent.)

Let $p:\mathbb{R}^2 \rightarrow T^2$ be the universal cover of $T^2$. Then since $f_*(\pi_1(S^2)) \subset p_*(\pi_1(\mathbb{R}^2))$ (both fundamental groups are zero) we get a lift $\tilde{f}:S^2\rightarrow \mathbb{R}^2$ such that $f = p\circ \tilde{f}$.

Since $H^2(\mathbb{R}^2) = 0$, $p^*:H^2(T^2)\rightarrow H^2(\mathbb{R}^2)$ must be the zero map.

Then $f^* = (p\circ \tilde{f})^* = p^*\circ\tilde{f}^* = 0\circ \tilde{f}^*$ is the zero map. Since $H^2(T^2) = H^2(S^2) = \mathbb{R}$ this contradicts that $f^*$ is an isomorphism. $\square$

Bohring
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  • (this is actually the definition used in the other problem but it is easy enough to see that the two are equivalent.)

    How's that equivalent?

     – xyzzyz Jan 17 '14 at 20:24
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    Related: http://math.stackexchange.com/questions/452916/map-of-arbitrary-degree-from-compact-oriented-manifold-into-sphere – Micah Jan 17 '14 at 20:26
  • @xyzzyz see http://www-personal.umich.edu/~wangzuoq/437W13/Notes/Lec%2031.pdf for example. – Bohring Jan 18 '14 at 00:47
  • @Pete there's nothing there about degree 1 maps inducing isomorphism on all cohomology groups. – xyzzyz Jan 18 '14 at 14:15
  • @xyzzyz Oooh - it was a typo on my part. They just induce isomorphisms on the top cohomology group. I fixed it now. Thanks for pointing this out. – Bohring Jan 18 '14 at 21:44
  • In this case, I don't get your argument about coverings and fundamental group -- it only proves something about $H^1$, and not about $H^2$ – xyzzyz Jan 19 '14 at 12:50
  • @xyzzyz I'm not quite sure what you mean... I added a few more details. – Bohring Jan 20 '14 at 09:58
  • Ah, I see your point. – xyzzyz Jan 20 '14 at 12:36

1 Answers1

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i) The natural projection $$ S^2\times S^3\to(S^2\times S^3)/(S^2\vee S^3)\cong S^5 $$ is such map.

ii) Suppose $f$ is a map $S^5\to S^2\times S^3$. Then $$ H^5(S^5)\ni f^*[S^2\times S^3]=f^*([S^2]\cdot[S^3])=f^*[S^2]\cdot f^*[S^3]=0. $$

Grigory M
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