Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $h$?

share|cite|improve this question
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
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.