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So a problem I have come across involves constructing some smooth degree 1 maps, and one such map to be constructed is a map $S^{2}\times S^{3} \rightarrow S^{5}$. I've been told that there is such a map as the projection $S^{2}\times S^{3} \rightarrow (S^{2}\times S^{3})/(S^{2} \lor S^{3})\cong S^{5}$, but I have no idea how to write out such a map in coordinates (to prove it is smooth) and I have some trouble proving that last isomorphism.

Many thanks for the help!

TheManWhoNeverSleeps
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  • You could look at the Sections of Chapter 5 of my book "Topology and Groupoids" which deal with the join and smash product of spaces. – Ronnie Brown Dec 28 '14 at 21:42
  • That book has some pictures. I'd be interested to know if it helps with the smooth case. – Ronnie Brown Dec 28 '14 at 21:52
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    I would, but chapter 5 is not available on your webpage, and as a poor student I would rather not order a book just to answer one small question :P, though the book does look pretty interesting. – TheManWhoNeverSleeps Dec 28 '14 at 21:59
  • Your library should get a copy! Also there is an e-version available for £5 - see my web page. – Ronnie Brown Dec 29 '14 at 10:35

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First of all, $S^2\times S^3$ has a CW structure consisting of one $0$-cell, one $2$-cell, one $3$-cell, and one $5$-cell. The subspace $S^2\vee S^3$ has a CW structure consisting of one $0$-cell, one $2$-cell, and one $3$-cell; these are just the corresponding cells of $S^2\times S^3$. Therefore, the space $S^2\times S^3/(S^2\vee S^3)$ has a CW structure consisting of one $0$-cell and one $5$-cell, so $S^2\times S^3/(S^2\vee S^3) \cong S^5$.

I assume you just need to find a smooth degree one map $S^2\times S^3 \to S^5$, you don't actually have to show that the given map is such a map. In that case, note that the given map $S^2\times S^3 \to S^5$ is a continuous degree one map, and is homotopic to a smooth map by the Whitney Approximation Theorem. As degree is preserved by homotopy, we're done.

Michael Albanese
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  • You mean if I just map the points $(x_{1}, x_{2})\times (x_{3}, x_{4}, x_{5})\rightarrow (1/2)(x_{1}, x_{2}, x_{3}, x_{4}, x_{5})$? This map is not surjective, for example the pole (1, 0, 0, 0, 0) is omitted. – TheManWhoNeverSleeps Dec 28 '14 at 22:53
  • No. I'm not sure if it is possible to write the map in local coordinates, but I don't know why you'd want to. Do you understand why $S^2\times S^3/(S^2\vee S^3) \cong S^5$? Once you understand that, you should see what the map is. – Michael Albanese Dec 28 '14 at 22:59
  • Well, it's just that I'm having trouble convincing myself that it is so intuitively obvious that the degree of the smash map is 1 that it is a detail I can just gloss over. Having the map in coordinates would allow me to actually see that is the case. Perhaps I am just being silly, and it really is just that obvious :P – TheManWhoNeverSleeps Dec 28 '14 at 23:19
  • @TheManWhoNeverSleeps The way one would generally show that a continuous map is degree 1 is to define degree homologically and verify it in this setting, which is probably what Michael is suggesting. –  Dec 28 '14 at 23:23
  • @TheManWhoNeverSleeps: Sorry, I didn't know that you were unsure of the degree. To 'see' that it has degree one, first note that the preimage of the $0$-cell of $S^5$ under the projection $S^2\times S^3\to S^5$ is a copy of $S^2\vee S^3$ inside of $S^2\times S^3$. The preimage of any other point under the projection is a single point. In a neighbourhood of this point (which does not intersect $S^2\vee S^3$), the map is an orientation-preserving diffeomorphism, so the degree is $1$. Maybe this is easier to see with the torus and the sphere. – Michael Albanese Dec 28 '14 at 23:33
  • Ooooh, I like @MikeMiller 's answer the best, but Michael's makes it easier to look at as well. :) – TheManWhoNeverSleeps Dec 28 '14 at 23:36
  • @TheManWhoNeverSleeps: In that case, you may be interested in this question. – Michael Albanese Dec 28 '14 at 23:39
  • @MichaelAlbanese That is truly lovely! – TheManWhoNeverSleeps Dec 28 '14 at 23:45