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Let $A,B\subset\mathbb{R}^{d}$ topological spaces and suppose that $B$ and $A\cap B$ are contractible. Do this conditions ensures that $A\cup B$ deformation retracts onto $A$ ?

So far, I haven't been able to find a counter example.

If this is not true, are there further conditions under which the statement is true?

BabaUtah
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1 Answers1

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A case where $A\cup B$ cannot even be continuously mapped onto $A$:

Let $A$ be the surface of a torus

Let $B$ be $C\cup C'\cup P$, where $C$ and $C'$ are disjoint open discs each of whose boundary is on $A$, and $P$ is a closed set which is the shortest path on $A$ connecting the boundaries of the circles.

EDIT: I meant that $C$ and $C'$ are meridians, look at the link if you need to find out what a meridian is:

Equators and meridians on a discrete torus

Chris Sanders
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  • Oh, yes, great counter example ! Thanks ! – BabaUtah Feb 03 '24 at 10:48
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    I think that those two circles need to be essential (e.g., a pair of longitudes or a pair of meridians). If we take two points of $A$, form small disks $D_1$ and $D_2$ around them, and push them off $A$ so they look like two igloos-without-entries, and call the interiors of the pushed-off-disks $B$ (along with the arc $P$) then there's a very natural map from $A \cup B$ to $A$, which is essentially "let each point of the igloo fall down to the point of the surface of $A$ just below it". – John Hughes Feb 03 '24 at 11:13
  • @JohnHughes Ah yes I meant two meridians – Chris Sanders Feb 03 '24 at 11:46