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2

In dimensions 2 and 3 every homeomorphism is isotopic to a diffeomorphism (this should be in Moise's book "Geometric topology in dimensions 2 and 3", it also follows from Kirby and Siebenmann's work). In dimension 4 there are self-homeomorphisms of simply-connected smooth compact manifolds which are not homotopic to diffeomorphisms. This follows e.g. from ...


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First, note that a homeomorphism $f$ of compact smooth manifolds is homotopic to a diffeomorphism if and only if one can approximate $f$ arbitrarily well by diffeomorphisms. For $n \leq 3$, this paper of Munkres claims as a corollary that a homeomorphism $f: M \to N$ of smooth manifolds may be approximated arbitrarily well by a diffeomorphism. This settles ...


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In this survey article on differential topology, Milnor outlines a proof that every PL manifold of dimension $n \leq 7$ possesses a compatible differential structure, and whenever $n<7$ this structure is unique up to isomorphism. He includes references for the various facts he uses.


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Let $(X_1,d_1),(X_2,d_2)$ be two geodesic metric spaces and $A$ be a proper metric space with two isometric embeddings $i_1 : A \to X_1$ and $i_2 : A \to X_2$. Finally, let $(X,d)$ denote the gluing $X_1 \coprod_A X_2$. We claim that $X$ is geodesic. Let $x_1 \in X_1$ and $x_2 \in X_2$. There exists a sequence $(a_n) \subset A$ such that ...


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Here are some thoughts about your question: Every topological space X which admits a complete CAT(0) metric is contractible, locally contractible and completely metrizable. In the locally compact case, such $X$ also admits a Z-compactification. I cannot think of any further restrictions on topology of such a space. One may conjecture that these are the ...


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Take the product of your favorite locally symmetric finite volume nonpositively curved complete Riemannian manifold with a sphere. For instance, take the product of $S^n$, $n\ge 2$, and a complete noncompact hyperbolic surface of finite area. The universal cover is a symmetric space. (In the specific example, it is $S^n\times H^2$.) Maybe you also want ...


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One thing you have to check ahead of time is connectivity; I'll presume that has been done. To determine orientability, do a depth first search to construct a polygon out of the 40 triangles. It will take 39 gluings. The resulting polygon will have 42 sides, glued in pairs. Check whether any of the 21 edge pair gluings reverse orientation. Alternatively, ...


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A necessary condition for an $n$-dimensional manifold $M$ to be immersible into $\mathbb{R}^{n+k}$ is that there must be a $k$-dimensional vector bundle $N$ on $M$ (the normal bundle) such that $T \oplus N$ is trivial, where $T$ denotes the tangent bundle. (By the Hirsch-Smale theorem, this necessary condition is actually sufficient when $k \ge 1$.) So we ...


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The above comments take two approaches, either use that (by definition) a bounded set is contained in a ball, or alternatively, that each coordinate is bounded. Assuming that a set $X$ is bounded (by definition) if there is $M>0$ such that $||x||\le M$ for each $x\in M$ (where $x=\{x_1,...,x_k\}$ and $||x||=\sqrt{x_1^2+...+x_k^2}$ ). Then $X$ is ...


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For Q2 $n=4$ is not possible. Let $c(\mathbf{x}) \in \{A,B,C,D\}$ denote the color of a point $\mathbf{x}$. As you noted in the comments, a 4-coloring with every unit square having four different-colored vertices would require C1: No pair of points separated by a distance of $1$ or $\sqrt 2$ can have the same color. From C1 we can show: C2: For any point ...


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The Jordan-Schoenflies theorem states that the inside and outside of a Jordan curve are homeomorphic to the inside and outside of a standard circle in $\mathbb{R}^2$. You can read more in this paper.


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Turning my comment into an answer: No, you cannot extend all homeomorphisms to the entire surface. For a counterexample, let $S$ be the surface of the unit sphere in $\mathbb{R}^3$ with the point $(0,1,0)$ removed. Let $R \subset S$ be the closed annulus consisting of points $(x,y,z)$ of $S$ with $\lvert y \rvert \leq \frac{1}{2}$. Then the complement of ...



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