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There are many more 3-manifolds that arise as the boundary of compact contractible 4-manifolds than just those, including some Brieskorn spheres like $\Sigma(2,3,5)$. The keyword you want is Mazur manifold. It is my impression that, and I would be surprised if it weren't true, there is not much known about what 3-manifolds arise as the boundary of a Mazur ...


1

A space is simply connected if any two points in the space can be joined by a continuous path lying in the space, and every continuous loop in the space can be continuously contracted to a point. For example, a circle fails to be simply connected, whereas a sphere is simply connected. Roughly speaking, a space is simply connected if there are no "tunnels" ...


2

Yes. You may use the splitting principle and assume that the vector bundle is a direct sum of $k$ line bundles, and note that the first Chern class of direct sum of line bundles is the sum of first Chern classes of the line bundles, which in turn is the Chern class of the tensor product of those bundles, which is isomorphic the top wedge power of the ...


0

I will try to extend on the idea I mention to solve the problem. So if $V$ is adjacent to edge $W_f$ and $W_g$, then we will give case by case relation between $f$ and $g$. As walls does not cross each other so $W_g$ is contained in $W_f^-$ of $W_f^+$. So a V is adjacent to $W_f$ and $W_g$ there is four cases assuming $f+g\neq 1$ Case 1: ...


5

The simplest possible example is the Klein bottle. One way to think of the Klein bottle is the quotient $S^1 \times I /\sim$, where $(z,0) \sim (\bar z,1)$ (considering $S^1 \subset \Bbb C$). (This way of defining it comes essentially from the standard representation; equating the red arrows gives you $S^1 \times I$, and the relation of equating the blue ...


2

Suppose your two knot exteriors are $M_1$ and $M_2$. van Kampen's theorem says that $\pi_1(M_1 \cup_h M_2)$ is the amalgamated free product of $\pi_1(M_1)$ and $\pi_1(M_2)$, amalgamated over $\pi_1(T^2)$, the fundamental group of their intersection. There is a canonical map $\pi_1(M_1) \to \pi_1(M_1 \cup_h M_2)$ given by inclusion $M_1 \hookrightarrow M_1 ...


3

First of all notice that $1-$dimensional manifolds generically embedded in $n-$manifolds are unlinked for $n\geq4$. As you already noticed your space $A := S^5 \setminus (S_1 \cup S_2 \cup S_3)$ is homeomorphic to $\mathbb{R}^5 \setminus (\mathbb{R} \cup S_2 \cup S_3)$. This space is homotopically equivalent to $(\mathbb{R}^5 \setminus \mathbb{R}) \vee ...



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