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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. $\square$

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  • $\begingroup$ > (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? $\endgroup$
    – xyzzyz
    Jan 17, 2014 at 20:24
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    $\begingroup$ Related: math.stackexchange.com/questions/452916/… $\endgroup$
    – Micah
    Jan 17, 2014 at 20:26
  • $\begingroup$ @xyzzyz see www-personal.umich.edu/~wangzuoq/437W13/Notes/Lec%2031.pdf for example. $\endgroup$
    – Bohring
    Jan 18, 2014 at 0:47
  • $\begingroup$ @Pete there's nothing there about degree 1 maps inducing isomorphism on all cohomology groups. $\endgroup$
    – xyzzyz
    Jan 18, 2014 at 14:15
  • $\begingroup$ @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. $\endgroup$
    – Bohring
    Jan 18, 2014 at 21:44

1 Answer 1

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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 $$ H^5(S^5)\ni f^*[S^2\times S^3]=f^*([S^2]\cdot[S^3])=f^*[S^2]\cdot f^*[S^3]=0. $$

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