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The result you want is false. Counterexamples are given in Yamamoto, Shuji and Yamashita, Atsushi, A counterexample related to topological sums. Proc. Amer. Math. Soc. 134 (2006), no. 12, 3715–3719 …

People have already mentioned the Bieberbach theorems, which imply that your manifold is homotopy equivalent to a Euclidean manifold. In fact, it is homeomorphic to the Euclidean manifold, at least i …

I think this can be extracted from Spanier's book. In Chapter 2.7, his Theorem 13 says that if B is a paracompact Hausdorff space, then a map p:E-->B is a fibration if and only if it is a local fibra …

I'm not sure about $\mathbb{R}^{2n}$, but you can embed them in $\mathbb{R}^{2n+1}$ using dimension theory. The theorem is that every compact metric space whose covering dimension is $n$ can be embed …

A weaker question replaces "homology class of an embedded submanifold" with $f_{\ast}([M])$ for some compact smooth manifold $M$ and an arbitrary continuous map $f:M \rightarrow X$. Once you give up …

The Lebesgue Number Lemma is absolutely not needed to compute $\pi_1(S^1)$, or more generally to compute $\pi_n(S^n)$. Here's one way to do it that I've actually used while teaching several times. Th …

There are two questions here. 1) The fact about Dehn twists and isotopies is really a consequence of the fact that the mapping class group of an orientable surface is generated by Dehn twists. For n …

I'm not sure exactly what you mean here. One possible interpretation to your questions is "Which discrete groups have classifying spaces that are smooth manifolds"? For this, here are a few isolated …

Yes. Observe first that $f$ can be first extended to an involution of $\mathbb{R}^3$ and then to an involution $F : S^3 \rightarrow S^3$ of the one-point compactification of $\mathbb{R}^3$. A classi …

In his paper MR0087106 (19,302f) Smale, Stephen A Vietoris mapping theorem for homotopy. Proc. Amer. Math. Soc. 8 (1957), 604–610. Smale proved the following theorem: Theorem : Let $X$ and $Y$ be …

It's not quite the same thing, but a related object is the Lyusternik–Schnirelmann category of a topological space. See http://en.wikipedia.org/wiki/Lyusternik-Schnirelmann_category

Assuming that your map $f\colon X \rightarrow Y$ is a map of CW complexes, the answer is yes. In fact, you can get away with quite a bit less. Assume that $X$ and $Y$ are arbitrary CW complexes equi …

$\newcommand{\Cone}{\operatorname{Cone}}$Let $d$ be the dimension of the manifold $M$. For $n \geq 2$, I will prove that the symmetric power $SP^n(M)$ is a manifold with boundary for $d=1$, a manifol …

$\newcommand\R{\mathbb{R}} \newcommand\cH{\mathcal{H}} \newcommand\bbD{\mathbb{D}}$ I'm marking this community wiki since it's not so much an answer but an attempt to make it clear to the casual reade …

For the Steenrod squares, I highly recommend the first couple of chapters of the book "Cohomology operations and applications in homotopy theory" by Mosher and Tangora. It's beautifully written (and …