Search Results

The set $X=\{0, \, 1\}$ endowed with the discrete topology shows that your question has a negative answer. In fact, the only non-constant self-map of $X$ different from the identity is the non-trivi …

Remark. This answer was written before the "infinite" assumption was added. Take the discrete space with two points. It is clearly homogeneous and Hausdorff, and its only self-maps are the two con …

It is known that the Cantor set minus a point is (up to homeomorphisms) the unique zero-dimensional separable metric space without isolated points that is locally compact and not compact. In particula …

According to this search in $\pi$-base, the sigma product of incountably many copies of $\mathbb{R}$ (no longer in pi-Base as of 2023 March) and the boolean product topology on $\mathbb{R}^{\omega}$ a …

Let $k$ be an algebraically closed field, ad take $X=\mathbb{P}^2_k$, $U=\mathbb{A}^2_k$. Then $X$ and $U$ are not homeomorphic, since $U$ contains two disjoint, Zariski-closed, irreducible subsets …

This is only a partial answer, however it is too long for a comment. What you want is actually true for compact subsets $K \subset \mathbb{R}^n$, i.e. in any dimension, under the additional assumptio …

Any real semialgebraic set $X \subset \mathbb{R}^N$ has a dense, open subset that is a submanifold: just take the complement of its singular set. In fact, the singular set is Zariski closed in $X$, he …

Denote by $N_c$ the non-orientable surface with $c$ pairwise disjoint crosscaps. So $N_1$ is the projective plane, $N_2$ is the Klein bottle, etc. Then the "reason" of Klein bottle's exceptionality i …

The case $g=0$ being straightforward, let me focus to the case $g \geq 1$. Let $X=\Sigma_g -\{p_1, \ldots p_k\}$ be the topological space obtained by removing $k$ distinct points from a closed surface …

A famous result in this spirit is the Ax-Grothendieck theorem, whose statement is the following: Theorem. If $f \colon \mathbb{C}^n \to \mathbb{C}^n$ is an injective polynomial function then $f$ …

Every compact topological manifold $M$ has the homotopy type of CW-complex. So, in the compact case, there is no such invariant of the type you are looking for (or, at least, such an invariant cannot …

As J.C. Ottem suggests, the best thing in those cases is to look for the original paper. Wikipedia says According to (Matoušek 2003, p. 25), the first historical mention of the statement of this t …

Edit. Peter Michor showed that the answer to the question is yes. By contrast, the following example shows that it is in general no when one replaces smooth path with regular path (I at first assumed …

This is well-known material, so insted of a "detailed answer" let me give you a standard reference. See B. Farb and D. Margalit: A primer on mapping class groups, Theorem 6.4.