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Dear Eric, here is a Bourbaki-style proof. Recall that a continuous map $f: Y\to X$ is called proper by Bourbaki if, for all spaces $Z$, the map $f\times 1_Z: Y \times Z\to X \times Z$ is closed. …

Dear trew, here is an application I like of Gel'fand-Naimark's representation theorem . I'm not sure it answers your question in a strict sense but I hope it is sufficiently close... . Consider a c …

No, there is no countably infinite compact Hausdorff topological group. Indeed such a group $G$ would have a left-invariant Haar measure $m$ with $m(G)=1$ and all points would have the same measure ( …

Yes, if $G$ is an abelian group, its classifying space $BG$ is an abelian topological group whose $\pi _1$ is $G$. You can find details in John Baez's wonderful post, hearteningly called "Classifyi …

Dear Michal, Munkres presents a regular space that is not completely regular as a very detailed exercise (more than half a page!) to §33 in his book "Topology, Second Edition, Prentice Hall,2000" (pag …

Consider the standard embedding of the unit interval in $\mathbb R^2$ viz. $I=[0,1]\times \{0\} \subset \mathbb R^2$. Let $C$ denote the Cantor subset $C \subset I$ and define $U= \mathbb R^2 - C$, an …

Dear Martin, your intuition is excellent. Here is a paper that indeed identifies closed subspaces of Hilbert spaces with the orthogonal projection onto them and thus studies the Grassmannian you are …

I nominate Ehresmann's theorem according to which a proper submersion between manifolds is automatically a locally trivial bundle. It is incredibly useful, in deformation theory for example, but is sa …

Every (?) algebraic geometer knows that concepts like homotopy groups or singular homology groups are irrelevant for schemes in their Zariski topology. Yet, I am curious about the following. Let's st …

Dear mustafa-kava, Amnon Neeman has written a rather down-to-earth book Algebraic and Analytic Geometry dedicated to a proof of Serre's celebrated GAGA theorem. Serre's article is 42 pages long and N …

0) I would guess that the compact spaces you are looking for are extremely rare. 1) For example the extremely simple contractible space $I=[0,1]$ is not suitable: Consider the inclusion $j\colon …

A Lie subgroup of a Lie group is usually defined as a subgroup which is also a submanifold. But actually any closed subgroup of a Lie group is automatically a manifold, hence a Lie subgroup. Similarly …

Claim: The complement $U=\mathbb C^h\setminus \{F=0\}$ is path-connected and thus connected. Proof: Given $a,b\in U$ consider the affine complex line $L_{a,b}=L$ joining $a$ to $b$. The polynomial $ …

It is easy to miss the point that in the second definition $\mathbb Q/\mathbb Z$ is required to be discrete in Hiro's question. Hence even if a continuous morphism $f:G\to T$ has image in $\mathbb …