# Tag Info

6

Take the loop space $\Omega(S^2)$.

6

One interesting example is the pseudocircle, which is the finite topological space $V=\{P_1,P_2,L_1,L_2\}$ where the open sets are those subsets $U$ such that if $P_1\in U$ or $P_2\in U$ then $L_1$ and $L_2$ are in $U$. My notation is suggestive; you can think of this space as a circle as follows Then the open sets are precisely the sets that 'look' open ...

6

To elaborate on Dmitri Pavlov's comment, your questions have been answered in a recent paper by Yehonatan Sella. Sella give an example (Example 0.3) of a locally contractible space $X$ for which $\mathcal{S}^k(X) \xrightarrow{\mathrm{shf}_X} \tilde{\mathcal{S}}^k(X)$ is not surjective. Sella's space has $5$ points, and the argument that $\mathrm{shf}_X$ is ...

6

Yes. Deleting a point from a manifold of dimension $n \geq 3$ doesn't change the fundamental group, so the result is still simply connected. (Either use van Kampen or transversality.) Mayer-Vietoris shows that the homology of the resulting space is the same as $S^2$, whence $\pi_2(U \setminus p)$ is isomorphic to $H_2(U \setminus p) = \Bbb Z$, generated by a ...

5

The quotient space $X = (\Bbb R \times \{0,1\}) / {\sim}$ where we identify $(x,0)\sim(x,1)$ for each $x\ne0$. This is called the real line with two origins, and it is not Hausdorff. We can cover $X$ with two open subsets $U$ and $V$, corresponding to $\Bbb R\times\{0\}$ and to $\Bbb R \times \{1\}$, respectively. Since their intersection $U\cap V$ has two ...

5

My favorite method of constructing a degree $1$ map is to take an embedded open ball in $M$ and crush its complement to a point.

4

I don't think it is true, what bout closed interval $[0,1]$, it is certainly a smooth manifold, and as you can prove it cannot have one $0-cell$. This result is true for closed smooth manifold, because you can find a Morse function on it with one minimum, i.e a critical point of index $0$. And then it corresponds to the $0-cell$ of the CW-presentation of ...

4

The ring $K(X)$ is never a field. If $X$ is empty, then $K(X)$ is the zero ring, which is not a field. If $X$ is nonempty, choose a point $x\in X$. Then there is a restriction homomorphism $r:K(X)\to K(\{x\})=\mathbb{Z}$. But there is no field that has a ring-homomorphism to $\mathbb{Z}$: if $p$ is any prime which does not divide the characteristic of a ...

4

Consider the space $K$ given by the disjoint union of two circles. If $\phi$ is the map interchanging the two components (preserving the orientations), then we see that $K_{\phi}$ is a torus. However, $K\times I$ is the disjoint union of two tori, hence the Betti numbers are obviously different: $$(b_0(K_\phi),b_1(K_\phi),\ldots)=(1,2,1,0,\ldots)$$ ...

4

If $f:M\to M$ is a diffeo, then there is a linear map $f^p:H^pM\to H^pM$ and then we can consider the spaces $$H^p(M)_f=H^p(M)/(1-f^p)H^p(M)$$ and $$H^p(M)^f=\ker(1-f^p).$$ One can then show that there are exact sequences $$0\to H^{p-1}(M)_f\to H^p(S_fM)\to H^p(M)^f\to 0$$ with $SM$ the mapping torus of $f$. One speedy way of getting this, by the way, is to ...

4

Definitely not. For instance, you could take your example $N\subset\mathbb{R}^3$ and form a connected sum $N\mathbin{\#} P$ where $P$ is any closed oriented 3-manifold. By some simple long exact sequence computations you can find that $H_2(N\mathbin{\#}P)\cong H_2(P)$, so $N\mathbin{\#}P$ cannot be homotopy equivalent to a wedge of circles if $H_2(P)$ is ...

3

Two examples of particular importance in symplectic topology are the Lie groups of unitary matrices $U(n)$ and symplectic matrices $Sp(2n, \mathbb{R})$. The symplectic matrices are the real $2n \times 2n$ matrices that satisfy $A^T J A = J$, where $$J = \begin{pmatrix} 0 & -I_n \\ I_n & 0 \end{pmatrix}$$ is the standard complex structure on ...

3

Look at the polygonal representation of two spaces. Now removing a disc from the middle, the rest of the space will deformation retract into the boundary, which is nothing but wedge of two circle. (Just draw the picture of polygonal presentation, you can actually see what is happening.)

3

Theorem:For a map $f:M\to N$ between connected closed oriented $n-$manifolds, suppose that there is a ball $B\subset N$ such that $F^{-1}(B)$ is the disjoint union of balls $B_i$ each mapping homeomorphically onto $B$. Then $deg(f)=\sum \epsilon_i$ where $\epsilon_i =\pm 1$ depending on whether $f|B_i$ preserves or reverses the orientation. You can prove ...

3

Your definition of $g_n$ doesn't seem right. If $z\in S^1$, i.e., $|z|=1$, then $nz\notin S^1$. To show that $(f_n)_*=\bullet \cdot n$, take your favorite generator $[\lambda]$ of $\pi_1(S^1)\cong \Bbb Z$. So $\lambda$ is a path that maps to $1$ in the fundamendal group. Then $(f_n)_*$ is determined by it's value on $[\lambda]$. But ...

3

Something which is generally true is that a homotopy equivalence $\alpha: X\to Y$ induces isomorphisms on all homotopy groups. That is, $\alpha_*: \pi_n(X) \to \pi_n(Y)$ is an isomorphism for all $n$. In the case $n=0$, $\pi_0$ is just a set, not a group, and $\alpha_*$ is a bijection. Thus $X$ is path connected iff $\#\pi_0(X) = 1 \iff \#\pi_0(Y) = 1$ iff ...

3

Given a $k$-manifold $M$, a necessary condition that it be the interior of a compact manifold with boundary is that there exists a compact submanifold with boundary $C \subset M$ such that the pair $(M-int(C),\partial C))$ is homeomorphic to the pair $(\partial C \times [0,\infty),\partial C \times 0)$. Perhaps the most well known counterexample is the ...

3

Take any (path connected) base-pointed space $X$ whose fundamental group has a subgroup isomorphic to $\mathbb{Z}$. Take a base-pointed topological space $Y$ and a covering map $Y \to X$ that corresponds to that subgroup, under the bijective correspondence between subgroups and pointed covering spaces. It follows that $\pi_1 Y \approx \mathbb{Z}$.

3

At most this requires the relative smoothing theorem: if $f: X \to N$ is $C^0$, and smooth on some closed subset $M$, then it is homotopic to a smooth map, with $f|_M$ fixed by this homotopy. Apply this to $X = S^n \times M$ to see that the map is a surjection on homotopy groups. (That is, we're smoothing a "sphere's worth" of continuous maps.) Now apply a ...

3

No, in any characteristic. Let $f:\mathbb{P}^1\times\mathbb{P}^1=X\to Y=\mathbb{P}^2$ be such a morphism. Then, the map can not be finite, since it is of degree one and $Y$ is smooth would imply that $f$ is an isomorphism. This is not true by Picard group considerations or many other considerations. So, one must have an irreducible curve $E\subset X$ such ...

3

Let $A$ be an affine space over the topological vector space $V$. I assume $A$ is topologized so that the action of $V$ on $A$ (given by $v \mapsto a_0+v$, $a_0$ some chosen point in $A$) is continuous. Then a contruction on $A$, given some choice of element $a_0 \in A$, is defined by $f_t(a) = a_0 + t(a-a_0)$. When one says '$A$ is affine over $V$' then ...

3

For $H_1$, there's a theorem that every finitely generated abelian group is a quotient of a finite rank free abelian group $\underbrace{\mathbb{Z} \oplus \cdots \oplus \mathbb{Z}}_{\text{$n$times}}$ for some $n$ (the theorem is actually more precise than this, but I'm writing it this way to emphasize $\mathbb{Z}$). For $\pi_1$, there's a theorem that every ...

3

No. As an obnoxious counterexample, take $g=0$ and $N$ to be the complement of some small open ball in a 3-manifold $M$; then $\pi_1(N) = \pi_1(M)$, so if $M$ was not simply connected (whence $S^3$, thanks to Perelman), then $N$ is not homotopy equivalent to the point. More generally, delete (a small neighborhood of) a wedge of $g$ circles from some ...

3

Closed surfaces are determined by the two properties: Orientability Euler characteristic As you have already correctly determined, the considered surface is not orientable, because the edge $c$ is glued with the opposite orientation. Thus one has to determine the Euler characteristic $\chi$, which can be calculated by the formula: \chi = \#\{Vertices\} ...

2

If your spaces are nice, then yes, Seifert-Van Kampen will show the wedge is simply connected. However, it's not true in general. Here's a counterexample due to Cannon and Conner ("On the fundamental groups of one dimensional spaces"). Take the Hawaiian earring space, $H$ and let $X=Y=C(H)$ be the cone on $H$. Now form a wedge sum $X\vee Y$ by identifying ...

2

I think you just forgot that $\mathbb{C}$ is two dimensional. Is $MU(n)$ in your notation the Thom space of the universal rank $n$ complex vector bundle on $BU(n)$? If so, because $\mathbb{C}$ has dimension two, the structure maps of the spectrum $MU$ go $\Sigma^2 MU(n) \to MU(n+1)$, which means that $MU(1) \mapsto \Omega^{\infty-2} MU$, not ...

2

The Möbius strip, or more generally anything that deformation retracts to $S^1$. For a less trivial example, take an infinite wedge of circles indexed by integers and for each triple of integers $x,y,z$ such that $x+y=z$, glue a 2-cell with boundary $x+y-z$. The fundamental group is generated by "1". (This correspond to the 2-skeleton of a common model of ...

2

$\newcommand{\Reals}{\mathbf{R}}$Answering the question as clarified in the comments: If $X$ is a topological space, and if $f_{1}$, $f_{2}:X \to \Reals^{3}$ are continuous, how can I define a homotopy between $f_{1}$ and $f_{2}$? Because $\Reals^{3}$ has operations of addition and scalar multiplication, you can get an infinite-dimensional family of ...

2

What I am going to write is essentially a simple rewrite of Hatcher's proof of van Kampen's theorem for your case, and without all the details (I can provide them if you want), which I hope it is clearer. Consider the homotopy $H \colon I \times I \to X$ of paths from $\gamma$ to $\theta$. By a Lebesgue number lemma argument, there are partitions $0 = s_0 ... 2 There is a homotopy$H_t: C_r \to \Bbb R^2 - 0$given by$H_t(z) = f(tz)$. At time$1$this is just$f(z)$and at time$0$it is the constant map$f(0)$. Note that$f(tz) \neq 0$by our assumption on$f(z)$. Therefore$f\$ induces the same homomorphism on homology as the constant map does, and the latter clearly induces the zero homomorphism. Edit: When I ...

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