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let $X$ be a topological space. suppose $\pi_i(X)=\mathbb Z$. let $f:S^i\rightarrow X$ be a representative of the generator of $\pi_i(X)$. $f$ induces an homomorphism $f_*:\pi_i(S^i)\rightarrow \pi_i(X)$ why $f_*$ is an isomorphism?

my guess: for every $\gamma:S^i\rightarrow S^i$

$f_*[\gamma]=[f\circ \gamma]$

this is injective as a homomorphism from $Z$ to $Z$ but why it is surjective? i mean take a class $[h]\in \pi_i(X)$ why would exist a map $\gamma:S^i\rightarrow S^i$ such that $f\circ \gamma$ is homotopic to $h$?

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Have you used the fact that $[f]$ generates $\pi_i(X)$? –  Jyrki Lahtonen Jun 7 '11 at 17:18
    
@ Jyrki Lahtonen : no because i don't see a way to manipulate $\alpha[f]$ for some $\alpha\in \mathbb Z$.. we can't write $[\alpha f]$ –  palio Jun 7 '11 at 17:37
    
Jim already showed how to use it (I wanted to give it as a hint) $f=f_*(id)$, $[id]$ generates $\pi_i(S^i)$ and $[f]$ generates $\pi_i(X)$... –  Jyrki Lahtonen Jun 7 '11 at 18:57
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up vote 3 down vote accepted

It's enough to check that the generator of $\pi_i(X)$ is hit by $f_*$. This follows because the generator of $\pi_i(S^i)$ is given by the identity map $id\colon S^i\to S^i$. So $f_*(id)=f$ is the generator you started with. Once you know the generator is hit, since $f_*$ is a homomorphism, this means that every multiple of the generator is also hit.

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