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  2. Retraction (topology) - Wikipedia

    en.wikipedia.org/wiki/Retraction_(topology)

    A space is an absolute neighborhood retract for the class , written ⁡ (), if is in and whenever is a closed subset of a space in , is a neighborhood retract of . Various classes C {\displaystyle {\mathcal {C}}} such as normal spaces have been considered in this definition, but the class M {\displaystyle {\mathcal {M}}} of metrizable spaces ...

  3. Karol Borsuk - Wikipedia

    en.wikipedia.org/wiki/Karol_Borsuk

    Karol Borsuk (8 May 1905 – 24 January 1982) was a Polish mathematician. His main area of interest was topology . He made significant contributions to shape theory , a term which he coined.

  4. Bing–Borsuk conjecture - Wikipedia

    en.wikipedia.org/wiki/Bing–Borsuk_conjecture

    In mathematics, the Bing–Borsuk conjecture states that every -dimensional homogeneous absolute neighborhood retract space is a topological manifold. The conjecture has been proved for dimensions 1 and 2, and it is known that the 3-dimensional version of the conjecture implies the Poincaré conjecture .

  5. Shape theory (mathematics) - Wikipedia

    en.wikipedia.org/wiki/Shape_theory_(mathematics)

    Shape theory is a branch of topology that provides a more global view of the topological spaces than homotopy theory. The two coincide on compacta dominated homotopically by finite polyhedra . Shape theory associates with the Čech homology theory while homotopy theory associates with the singular homology theory.

  6. Homotopy extension property - Wikipedia

    en.wikipedia.org/wiki/Homotopy_extension_property

    The homotopy extension property is depicted in the following diagram If the above diagram (without the dashed map) commutes (this is equivalent to the conditions above), then pair (X,A) has the homotopy extension property if there exists a map f ~ ∙ {\displaystyle {\tilde {f}}_{\bullet }} which makes the diagram commute.

  7. Borsuk's conjecture - Wikipedia

    en.wikipedia.org/wiki/Borsuk's_conjecture

    [7] For all n for fields of revolution — shown by Boris Dekster (1995). [8] The problem was finally solved in 1993 by Jeff Kahn and Gil Kalai, who showed that the general answer to Borsuk's question is no. [9] They claim that their construction shows that n + 1 pieces do not suffice for n = 1325 and for each n > 2014.

  8. Using the Borsuk–Ulam Theorem - Wikipedia

    en.wikipedia.org/wiki/Using_the_Borsuk–Ulam...

    Using the Borsuk–Ulam Theorem: Lectures on Topological Methods in Combinatorics and Geometry is a graduate-level mathematics textbook in topological combinatorics. It describes the use of results in topology , and in particular the Borsuk–Ulam theorem , to prove theorems in combinatorics and discrete geometry .

  9. Borsuk–Ulam theorem - Wikipedia

    en.wikipedia.org/wiki/Borsuk–Ulam_theorem

    The Borsuk–Ulam theorem has several equivalent statements in terms of odd functions. Recall that S n {\displaystyle S^{n}} is the n -sphere and B n {\displaystyle B^{n}} is the n -ball : If g : S n → R n {\displaystyle g:S^{n}\to \mathbb {R} ^{n}} is a continuous odd function, then there exists an x ∈ S n {\displaystyle x\in S^{n}} such ...