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Fundamenta Mathematicae.

Biblioteka Wirtualna Matematyki.

European Digital Mathematics Library.


Jul
7
asked Computing the fundamental group of a cell complex
Jul
7
accepted Quotient of free group with normal subgroup
Jul
7
asked Quotient of free group with normal subgroup
Jul
5
comment Degree of an odd map
oh noes, that's not possible because as you pointed out in your answer: the endpoints are antipodal. Maybe it's time for me to take a coffee break now : (
Jul
5
accepted Degree of an odd map
Jul
5
comment Degree of an odd map
thank you! And why does it follow that $f \circ \gamma_+$ also sweeps an angle of $\pi + 2 \pi n$? Could it not be that it sweeps an angle $\Theta + 2 \pi n$?
Jul
5
comment Degree of an odd map
My goal is to use this to show that an odd map $f:S^1 \rightarrow S^1$ has odd degree...
Jul
5
comment Degree of an odd map
I am meaning to use $S^1 = [0,1]/\{0, 1\}$!
Jul
5
comment Degree of a retraction
You're not missing anything! Thank you! Maybe you could post this as an answer, I'll accept it.
Jul
5
comment Degree of a retraction
hm... no, I actually meant $r: S^1 \rightarrow S^1$. Why consider $r \circ i$ instead of $r$? If $r: S^1 \rightarrow S^1$ there has to be a way to find $deg(r)$.
Jul
5
asked Degree of an odd map
Jul
4
accepted Degree of a retraction
Jul
4
comment Degree of a retraction
Oh well : ) Ok, I assume no comment means yes what I repeated is what you meant. Thanks for your help!
Jul
4
comment Degree of a retraction
I think I have to add '@....' in order to appear in your inbox.
Jul
4
comment Degree of a retraction
Thank you!! Let me repeat to see if I understand correctly: $deg (i) = 1$ because the inclusion is the identity. $deg (i \circ r) = deg (r \circ i) = 1$ because both are homotopic to the identity functions (respectively). Therefore $deg (i \circ r) = deg(r) deg(i) = deg(r) \cdot 1 = 1$ implies $deg(r) = 1$.
Jul
4
comment Degree of a retraction
@omar: no, I'm interested in the case $f(z) = \frac{z}{|z|}$.
Jul
4
revised Degree of a retraction
Clarification added in response to comments.
Jul
4
asked Degree of a retraction
Jun
30
comment A question regarding the proof of $\pi_1(\mathbb{R}^2 \backslash \{ 0 \}, 1) \cong \mathbb{Z}$
Thanks, that seems much shorter. Why would anyone do something complicated like in the link above?
Jun
30
accepted A question regarding the proof of $\pi_1(\mathbb{R}^2 \backslash \{ 0 \}, 1) \cong \mathbb{Z}$