Let $A_n=\{z\in \mathbb{C}\mid z^n$ is non-negative real number$\}$ then find
$H_1(A_n,A_n-\{0\})$
$H_1(A_n,A_n-\{z\})$ when $0\not=z\in A_n$
show that $A_n$ is not homeomorphic to $A_m$ when $n\not= m$
I can only show that when $m\not=n$ is some small number(for example n=2,m=3) that $A_2$ $A_3$ and $A_4$ are not homotopy equivalent by just counting the number of connected component when removed some points.However,since I didn't catch up with the class that has finished studying about Van Kampen and cell-complex,I cannot do this using homology knowledge.
Can anyone explain how to do this problem?
your answers would be really appreciated.
Here's a hint which should be sufficient.
Suppose $A_n$ and $A_m$ are homeomorphic and the homeomorphism is given by $f\colon A_n\to A_m$. It's not hard to see that there must then be a homeomorphism between the subspaces $A_n\setminus\{0\}$ and $A_m\setminus\{f(0)\}$ given by a function $ f_0 \colon A_n\setminus\{0\}\to A_m\setminus\{f(0)\}$ defined by $f_0(x)=f(x)$. How many path components does $A_n\setminus\{0\}$ have?
I should say that this doesn't quite work for $n=2$ or $m=1$. In such cases, you can instead consider $g\colon A_m\to A_n$.
It's likely that $H_1(A_n,A_n\setminus\{0\})$ and $H_1(A_m,A_m\setminus\{f(0)\})$ are not isomorphic and $H_1(A_n,A_n\setminus\{0\})$ and $H_1(A_m,A_m\setminus\{0\})$ are not isomorphic for all $n\neq m$, but the above 'cut-point' argument is more elementary (if your lecturer/teacher wants you to use homotopy methods, then you should use the above homology argument).