I am trying to show that $g^*$ is non-trivial, but I am having a hard time justifying it based on this commutative diagram. Is $g^*$ just an isomorphism? Here is a screenshot of my work. I'd appreciate any feedback.
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1Here is a related post: link – Kevin.S May 11 '21 at 09:02
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2The conclusion at the end is not right: even if $h$ is an isomorphism, $g_$ could be trivial, and hence $g^$ would be trivial too. For example, $g$ could be a constant map. Perhaps what you want to say is that if $g_$ is not trivial then $g^$ is not either. – Pedro May 11 '21 at 12:03
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Ah, yes. That rings true. Thank you, Pedro! – blueskyscroll May 11 '21 at 12:48
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1If $g$ is the constant map, then $g^*$ is trivial. – Paul Frost May 11 '21 at 17:01
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Think of $T^2$ as a square with opposite sides identified. Then the map $g: T^2\to S^2$ induced by collapsing the four sides of the square to a single point induces an isomorphism $g^*: H^2(S^2, \mathbb{Z})\to H^2(T^2, \mathbb{Z})$.
Alex Fok
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