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I was going through a paper by Tanaka where I am stuck at the following enter image description here map "f" which is given by the composition of these maps. Next, he mentions that the induced map is clearly an isomorphism. Kindly, help me to figure out why is that true.

I think I need to prove that all individual maps induce an isomorphism. Obviously, the inclusion induces an isomorphism so does the homeomorphism '$h$'. What about the maps $j$ and $q$?

Any hints will be of great help!!

Thanks in advance.

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    $\begingroup$ It might help me a bit to see what $\xi$ is, and I don't know for sure what is meant by the superscript $d$ on the upper right. Is it meant to be $d$-skeleton of the suspension? That of the projective space itself is the whole space. $\endgroup$
    – jdc
    Commented Oct 30, 2022 at 12:54
  • $\begingroup$ Thanks @jdc, for your response. $\xi$ is the canonical bundle over that projective space and that notation $d\xi$ means the Whitney sum of d copies of $\xi$, only $d$ represents trivial bundle of rank $d$ & when we take it as a superscript it represents the Thom Space of that bundle. $\endgroup$
    – Devendra Singh Rana
    Commented Oct 30, 2022 at 15:48
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    $\begingroup$ For $q$, it is the $d$-fold suspension of a map $(\mathbb R P^{d-1})^d \to {*}^d = S^d$. By the naturality of the Thom isomorphism, the map induced on $H^d$ by this map will be an isomorphism if the map induced on $H^0$ by the collapse map $\mathbb R P^{d-1} \to {*}$, and we do know this. What is $\mathbb R P^{d-1}_d$? $\endgroup$
    – jdc
    Commented Oct 30, 2022 at 22:45
  • $\begingroup$ That actually a single point, right? @jdc $\endgroup$
    – Devendra Singh Rana
    Commented Oct 31, 2022 at 4:23
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    $\begingroup$ I'm not sure... I was trying to ask you what the notation meant. $\endgroup$
    – jdc
    Commented Oct 31, 2022 at 12:09

1 Answer 1

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As explained in comments by jdc $q$ induces an isomorphism.

That leaves us with $j$.

By taking a look at the cohomology of $RP^{2d-1}$ and its $(d-1)$-st skelton (which becomes $(2d-1)$-st skelton after suspending d times), which is just $RP^{d-1}$ we also see that $j$ induces an isomorphism in $H^{2d}$.

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