If I have a square and I remove an open disc from its interior, there exists a deformation retraction onto its boundary. Is this also the case, if I remove a closed disc from its interior? Does the same retraction work?
2026-04-29 07:41:39.1777448499
Deformation retraction onto the boundary
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Yes, the retraction is given by mapping the point $x$ to the intersection of the boundary of the square and the line which goes through $x$ and the center of the removed disk (open or closed). As the center is in the interior of the square, this is well defined and, if we move $x$ along the line uniformly as $t$ runs from $0$ to $1$, then this is also a deformation retract.