I want to prove what is stated from above. I feel like it should just follow from the definition of the fundamental group, but I'm not entirely sure where to go from there (sorry if I'm missing something obvious!).
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1$\pi_2(B)$? The second homotopy group of $B$? – Angina Seng Nov 05 '17 at 19:16
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1I presume you mean $\pi_1(B)$? – Clive Newstead Nov 05 '17 at 19:16
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For the converse see this question. – Dietrich Burde Nov 05 '17 at 19:19
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Yes, sorry, I made an error, it was supposed to be $\pi_2$. I have edited it now, sorry for the confusion. – Nov 05 '17 at 19:23
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You really need $A$ and $B$ to be path-connected for this question to make sense. Finicky topologists would write $\pi_1(A,a)$ rather than $\pi_1(A)$ where $a\in A$ is a base-point. – Angina Seng Nov 05 '17 at 19:31
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Use functoriality: functoriality tells us that $h_$ is an isomorphism, since $id_=(hh^{-1})*=hh^{-1}_$. – Andres Mejia Nov 05 '17 at 19:37
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Possible duplicate of Homeomorphic spaces have the same homology groups – Harambe Nov 06 '17 at 06:04
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Let $f : A \to B$ be a continuous map. If $u : [0,1] \to A$ is a loop in $A$, then $f \circ u : [0,1] \to B$ is a loop in $B$. This descends to a map of homotopy groups $$f_* : \pi_1(A) \to \pi_1(B)$$ defined by $f_*([u]) = [f \circ u]$ for all homotopy classes $[u]$ of loops in $A$.
You should check that $f_*$ is a well-defined group homomorphism, and that when $f$ is a homeomorphism, the map $f_*$ is an isomorphism of groups.
Clive Newstead
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