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This question was prompted by my answer to this question. An exotic $\mathbb{R}^4$ is a smooth manifold homeomorphic to $\mathbb{R}^4$ which is not diffeomorphic to $\mathbb{R}^4$ with its standard s …

Let $M$ be a smooth $n$-manifold. Its tangent bundle $TM$ is diffeomorphic to $\mathbb{R}^{2n}$ if and only if $M$ is open and contractible. Proof: One direction is straightforward. Any vector bundl …

Is there a closed, smooth, orientable manifold which is not spin${}^c$ but has a finite cover which is spin${}^c$? Such examples exist when spin${}^c$ is replaced by spin: an Enriques surface is …

Consider $M = \mathbb{CP}^2\#\mathbb{CP}^2$. All of its Stiefel-Whitney numbers vanish, so $M$ is unorientedly nullcobordant; more generally, $X\# X$ always bounds. We have $H^2(M; \mathbb{Z}) \cong …

Recall that on a closed $n$-manifold $M$, there is a unique class $\nu_k$ such that $\operatorname{Sq}^k(x) = \nu_kx$ for all $x \in H^{n-k}(M; \mathbb{Z}_2)$; this is called the $k^{\text{th}}$ Wu cl …

The manifolds $M$ and $N$ may not even be homotopy equivalent! In Compact Flat Riemannian Manifolds: I, Charlap showed that there are two closed flat manifolds $M$ and $N$ of the same dimension which …

Let $X$ be a smooth, compact, orientable manifold and let $\omega$ be a choice of volume form. On $X\times X$, we have the natural volume form $\sigma = \pi_1^*\omega \wedge \pi_2^*\omega$ where $\pi_ …

Since $\pi\circ\iota$ is constant, we have $d\pi\circ d\iota = 0$, so the image of $d\iota$ is contained in the kernel of $d\pi$, i.e. $(d\iota)(T\mathbb{CP}^1) \subseteq L_n$. Since $\iota$ is an emb …

A homotopy sphere is a topological $n$-manifold $M$ which is homotopy equivalent to $S^n$. A homology sphere is a topological $n$-manifold $M$ such that $H_i(M) \cong H_i(S^n)$ for all $i$. Note, by …

As discussed in this question from last week, if $M$ is a closed manifold such that $M\times S^1$ is homeomorphic to the torus $T^{n+1}$, then $M$ is homeomorphic to $T^n$. Is the corresponding statem …

Thanks to Igor Belegradek for his help in the comments. This answer is merely an expansion of his remarks. The proof of the following proposition was adapted from the paper A simplification problem in …

If $M$ is an $n$-dimensional complex manifold, then $M$ is a $2n$-dimensional smooth manifold, so you should integrate a $2n$-form (note that $\bar{\partial}$ of a $(0, n-1)$-form is an $n$-form). C …

Here's an elementary observation that gives a relation between the three sets in both cases. Note that $M_1\# M_2$ admits degree one maps $\alpha : M_1\# M_2 \to M_1$ and $\beta : M_1\#M_2 \to M_2$ ( …

Here's an example to show that the answer to the first question is no. Let $M = \mathbb{CP}^2$ and $N = \mathbb{CP}^1$. Note that $M - N$ is diffeomorphic to $\mathbb{C}^2$ and is therefore contractib …

I don't know the answer for general dimensions, but here is an argument that shows that the four-dimensional manifolds suggested by valeri fit your criteria (provided $n > 0$). The Dold-Whitney Theor …