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Someone
  • Member for 3 years, 11 months
  • Last seen more than 1 year ago
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awarded
 Yearling
awarded
 Curious
comment
Degree one self-map of $\Bbb R^2\big\backslash \big\{(n,0):n\in \Bbb Z\big\}$ not homotopic to any self-homotopy equivalence
I can not accept both nice answers(similar). So, I decided to do this: for one case, one upvote but not ✔, and for another one, ✔ but not upvote. Sorry.
comment
Degree one self-map of $\Bbb R^2\big\backslash \big\{(n,0):n\in \Bbb Z\big\}$ not homotopic to any self-homotopy equivalence
I can not accept both nice answers(similar). So, I decided to do this: for one case, one upvote but not ✔, and for another one, ✔ but not upvote. Sorry.
accepted
Degree one self-map of $\Bbb R^2\big\backslash \big\{(n,0):n\in \Bbb Z\big\}$ not homotopic to any self-homotopy equivalence
comment
Degree one self-map of $\Bbb R^2\big\backslash \big\{(n,0):n\in \Bbb Z\big\}$ not homotopic to any self-homotopy equivalence
Everywhere, compact inverse image is compact. I am taking this as a definition. Just as a note, all my surfaces are boundaryless, so there is no chance of considering properness in terms of the boundary.
asked
Degree one self-map of $\Bbb R^2\big\backslash \big\{(n,0):n\in \Bbb Z\big\}$ not homotopic to any self-homotopy equivalence
asked
Uniqueness of collar neighborhoods for non-compact boundary case in smooth setting
revised
Codimension one submanifold gives cofibrant pair
edited tags
asked
Codimension one submanifold gives cofibrant pair
awarded
 Disciplined
awarded
 Cleanup
awarded
 Autobiographer
awarded
 Scholar
awarded
 Supporter
accepted
Lifting of a proper map in the cover is a proper map
awarded
 Editor
revised
Lifting of a proper map in the cover is a proper map
added 1 character in body
awarded
 Student
asked
Lifting of a proper map in the cover is a proper map