Isomorphism between cohomology groups

Question

Let $(S^3, *), (S^5, *)\in Top_*$ be two spheres. We have $S^3 \vee S^5 = S^3 \times \{*\} \cup \{*\} \times S^5.$ The project map $\pi: S^3 \vee S^5\to S^3$ a continuous. Hence $\pi$ reduced to cohomology group isomorphism $$\pi^*: \mathbb{Z}/2\cong H^3(S^3; \mathbb{Z}/2) \to H^3(S^3\vee S^5; \mathbb{Z}/2)\cong \mathbb{Z}/2.$$

My question is why $\pi^*$ an isomorphism??

I hope that someone, can help. Thank you very much!

score 4 · Answer 1 · answered Dec 06 '16 at 05:59

4

Consider the obvious map $q:S^3\to S^3\vee S^5$. The composition $\pi\circ q$ is the identity map of $S^3$. Can you use this for something?

answered Dec 06 '16 at 05:59

Mariano Suárez-Álvarez

139,300

@ Mariano Suárez-Álvarez. We have $id = (\pi\circ q)^* = q^\circ \pi^$ and $id = (q\circ \pi)^* = \pi^\circ q^ ,$ so $\pi^*$ an isomorphism.right or wrong? – lony Dec 06 '16 at 06:14
One of the two compositions is not the identity. – Mariano Suárez-Álvarez Dec 06 '16 at 06:15
@ Mariano Suárez-Álvarez. Yes! How to prove $\pi^*$ an isomorphism? – lony Dec 06 '16 at 06:18
Well, use my hint. – Mariano Suárez-Álvarez Dec 06 '16 at 06:19
@ Mariano Suárez-Álvarez. I don't understand. Can you explain for me? – lony Dec 06 '16 at 06:21
2

I suggest you think about this for a longer time. There is no hurry. – Mariano Suárez-Álvarez Dec 06 '16 at 06:23

Isomorphism between cohomology groups

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