New answers tagged

2

No, having a CW complex structure is absolutely not enough to be a manifold. Being a CW complex is very easy, being a manifold is hard. Take the wedge sum of two circles $S^1 \vee S^1$ for example: it's a CW complex, but not a manifold. If a space $M$ is a compact manifold without boundary, then you can read its dimension using singular homology: it is the ...


2

This follows from the standard result given in, for example, Topology and Groupoids: $4.3.2$ Let $f : X \to Y$ be an identification map and let $B$ be locally compact. Then $$f × 1 : X × B \to Y × B$$ is an identification map. The definition of locally compact here is that each point has a base of compact neighbourhoods. This inconvenient ...


1

$\newcommand{\Z}{\mathbb{Z}}$ The question is equivalent to when $X$ can be written as a product of eilenberg mclane spaces. First why the question is important: The motivation for introducing Postnikov towers is to show that every space can be written as a fibered product of Eilenberg MacLane spaces. It is desirable to know if the $k$ invariants, the ...


2

This is carried out in a lot of detail in Voisin – Hodge theory and complex algebraic geometry I, section 11.1.2. The particular result you're interested in seems to be Corollary 11.15.


1

I think this is related (or maybe follows from) the result by E. Michael that the product of a locally compact space and a $k$-space is again a $k$-space.


2

All of the homology groups of $\mathbb{S}^n$ are trivial, except of top and bottom one. The induced map $H_0(\mathbb{S}^n) \to H_0(\mathbb{R}\mathbb{P}^n)$ will always be isomorphism (this is very easy to calulate). The top homology group $H_n(\mathbb{R}\mathbb{P}^n)$ is either trivial or $\mathbb{Z}$, depending on whether $\mathbb{R}\mathbb{P}^n$ is ...


2

Yes, it is possible, and so wedge sum for finite CW complexes does not satisfy homotopy cancellation. There is a description of some examples in the article "On principal $S^3$-bundles over spheres" by P. Hilton and J. Roitberg, although they mention they already appear in "On the Grothendieck group of compact polyhedra" by P. Hilton. Here is the idea of the ...


4

Are you familiar with this picture? (courtesy of Wikimedia) To get a torus you start with a full square $[0,1]^2$ and then you identify opposite edges in the way depicted above. Indeed if you first identify the red edges you get a cylinder, and then the blue edges become the top and the bottom circle. If $x$ is for blue and $y$ is for red, the ...


4

No, as it does not reflect products. For example the cone $\mathbb{R} \gets \mathbb{R} \to \mathbb{R}$ over the constant diagram $\{ \mathbb{R} \quad \mathbb{R} \}$ gets sent via the localization functor to a cone isomorphic to $* \gets * \to *$ (where $*$ is a singleton). This is a product in $\mathsf{hTop}_*$ (a singleton times a singleton is a singleton). ...


1

Here are some thoughts that could hopefully help. First of all, trying to give the product of two simplices a simplicial (or $\Delta$-complex) structure is in general annoying. There's a reason that the relevant sections of Hatcher are rather technical (see for example the proof of Theorem 2.10 - page 112 here (PDF)). Here's one thing that is neat about ...


1

You will have to get used to the fact that sometimes pictures suffice to give a proof. The idea is that a picture can be made "rigorous" by writing down an explit homeomorphism from the picture. A lemma that is often implicitly used here is the following: Let $Y$ be a Hausdorff space, and let $X$ be a compact, and $\sim$ and equivalence relation on ...


10

As soon as you have a homotopy invariant functor, it factors through the homotopy category, by the universal property of localizations. To give a homotopy invariant functor $\mathsf{C} \to \text{Whatever}$ is exactly the same thing as giving a functor $\operatorname{Ho}(\mathsf{C}) \to \text{Whatever}$. So if we can understand the homotopy category well ...


0

The following theorem is given with proof here: Let $X$ be a topological space that is path-connected and locally path-connected. Let $G < \text{Hom}(X)$ be a group of homeomorphisms on $X$. Then the projection map $q: X \to X/G$ is a covering map if and only if $G$ acts properly discontinuously on $X$. Properly discontinuously means that for every $x ...


3

Some obvious things are less obvious than others.... Let $X$ be a rank two graph consisting of a disjoint pair of circles plus an arc with one endpoint on each circle. Then no matter what $v \in X$ you pick, $\pi_1(X,v)$ is not generated by simple closed curves through $v$.


3

By default, when you talk about representable functors they have to be functors to $\text{Set}$. If you want them to take values in a more refined category that can sometimes happen if the representing object itself has additional structure. Here $S^1$ has the structure of a cogroup (in the pointed homotopy category), which corresponds to the group ...


0

The standard reference is de Rham's Differentiable Manifolds, which develops the whole theory in terms of currents. But more quickly, see the beginning of Section 7.3 of Nicolaescu's notes http://www3.nd.edu/~lnicolae/Lectures.pdf A current is a functional, so it must eat differential forms and spew out numbers. This integral $\int_X a\wedge j_e$ is nothing ...


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\} ...


0

The notation in your picture is inconsistently overdetermined, and this seems to be causing you confusion. The arrows on the sides are already enough to indicate orientation, so you should not also put a superscript $-1$ on any arrow labels. Thus you should replace the $a^{-1}$ by an $a$ and the $b^{-1}$ by a $b$. It is then clearer that you can combine ...


2

The following proof is a bit over the top, but here it is: Prove Gauss' isothermal coordinates theorem (every 2-dimensional Riemannian manifold is conformally flat). As a corollary, conclude that every oriented 2-dimensional Riemannian manifold $(M,g)$ has a natural structure of a Riemann surface (the holomorphic atlas is given by charts $\phi: U\subset ...


6

It is true that if you have a homotopy $F : X \times [0,1] \to Y$ between $f,g : X \to Y$, then for all $x \in X$ this defines a path between $f(x)$ and $g(x)$, given by $t \mapsto f(x,t)$. This is a "if... then..." statement. The converse need to be true. If I give you a map $F : X \times [0,1] \to Y$, and I tell you that for all $x$, $t \mapsto F(x,t)$ is ...


3

Let us recall the formal definition of a homotopy between $f$ and $g$. Suppose $f,g : X\to Y$ are continuous functions, then a homotopy $H: X \times I \to Y$ is a continuous function such that $H(x,0) = f(x)$ and $H(x,1) = g(x)$ for all $x\in X$. What's the difference between this and what you've mentioned? Your condition isn't enough to satisfy that we get ...


2

As others have said, an augmented chain complex is an $(\mathbb{N} \cup \{-1\})$-graded chain complex $C_*$ with $C_{-1} = \mathbb{Z}$ (or more generally your base ring $\Bbbk$) and such that $H_{-1}(C) = 0$, equivalently the map $C_0 \to C_{-1} = \mathbb{Z}$ is surjective. The splitting result follows from the projective property of $\mathbb{Z}$ among ...


2

One considers $\mathbf Z$ as a complex, the complex all of whose chain groups are $0$ except the one in degree $0$ where we have the group $\mathbf Z$. We write this complex as $\mathbf Z[0]$. An augmentation of a complex $C$ is then an epimorphism of complexes $C\rightarrow \mathbf Z[0]$. A morphism of two augmented chain complexes $\delta\colon ...


1

From some googling, apparently an "augmented chain complex" is defined to be a chain complex together with a surjective homomorphism $C_0\to\mathbb{Z}$ such that the composition $C_1\to C_0\to\mathbb{Z}$ is $0$. This meaning seems compatible with your homework problem. I must confess that I had never seen this definition before now; if someone asked me ...


0

From a short exact sequence $$0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0,$$ which is easier than the case you describe, knowing $A$ and $B$ in general is not enough to compute $C$. This is the extension problem. Some examples $$ 0\rightarrow \mathbb{Z}\rightarrow \mathbb{Z}\rightarrow 0\rightarrow 0, $$ or $$ 0\rightarrow ...


0

Let $\pi \colon M \to S^1$ denote the usual projection and let $C$ denote a non-zero homologous simple closed curve with no self intersections, i.e. an injective map $c$ from $S^1$ to $M$, which induces something non-zero on homology. We can isotop $c$ s.th. $\pi \circ c$ is a covering of $S^1$ (This is not that hard, you just make the map locally ...


0

Every ball $B(0,r)$ is contractible, via the contraction $x \to tx$. This gives that every two maps from any space into the ball are homotopic, as a contractible space has the homotopy type of a point. If you think about isotopies and embedded then the story changes completly and depends on the dimension of the ball. In dimension greater than $4$ and $2$ ...


5

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 ...


4

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

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 ...


1

I interpret your question as follows: "Is there a (nicely) embedded graph in $S^3$ whose complement is homotopy equivalent to the projective plane or Klein bottle?" To see why both of your examples fall under this, think of $S^3$ as $\Bbb R^3$ with a point at infinity. Then your first example is the complement of two circles in $S^3$ (the picture you get is ...


1

1) You are correct. In the case that there is a covering map $p:\widetilde{X} \to X$, $\widetilde{X}_\rho$ is defined to be the quotient of $\widetilde{X}_0\times F$ by identifying all pairs $([\gamma],\widetilde{x}_0)$ and $([\lambda \cdot \gamma],L_\lambda(\widetilde{x}_0))$ for all $[\lambda] \in \pi_1(X,x_0)$. The only property of the covering space used ...


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

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: $$(b_0(K_\phi),b_1(K_\phi),\ldots)=(1,2,1,0,\ldots)$$ ...


0

It suffices to show the orbit $G(0)$ of $0$ is a discrete set, and any $x\in\mathbb{R}$ is "sandwiched" by two representatives of $G(0)$. Then the space $X/G$ is homeomorphic to $\mathbb{S}^1$ (the action is continuous), and so $G\approx\pi_1(X/G)=\pi_1(S^1)=\mathbb{Z}$. The discreteness is natural. Suppose $G(0)$ is bounded above. Let $s\equiv\sup G(0)$; ...


2

I'm confused. $H_1(\mathbb{R}P^3)=\mathbb{Z}/{2\mathbb{Z}}$, but is orientable (on edit, this answers your question 2). EDIT: Maybe it refers to the fact that $H^{m-1}(M)$ can only be torsion if the manifold $M$ is non-orientable. It is Corollary 3.28 in Hatcher.


1

a) Cohomology should be regarded as a graded-commutative algebra, therefore is a finer invariant. This allows for some quite interesting definitions, e.g. the Hopf invariant. b) If $M$ is a manifold of suitable dimension, then one has $b_{n-k}=b_k$ for the Betti numbers (proved via Poincare duality, which involves cohomology). It is sometimes easier to ...


2

There are papers on equivariant crossed complexes, see [93,114] on my publication list, where the second deals with the topological group case. A second possibly related question is to deal with Galois cohomology, using crossed complex methods. There are papers of M.V.Borovoi on nonabelian Galois cohomology with coefficients in a crossed module.


7

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 ...


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 ...


1

To answer the implicit question about the equivalence of sheaf cohomology and singular cohomology: local contractibility, in fact, semi-local contractibility, is sufficient to establish the equivalence, see the paper of Sella (Comparison of sheaf cohomology and singular cohomology, arXiv:1602.06674v3), where he also explains how Ramanan uses the hereditary ...


1

$\pi_1(A)$ and $\pi_1(B)$ are trivial since they are contractible, $\pi_1(C)$ is the free group generated by 2 elements since $C$ retract to the bouquet of two loops.


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 ...


1

(1). The def'n of the topology $T_d$ generated by a metric $d$ on a set $S$ is the topology whose base is the set of open $d$-balls. (I dk why you don't want to use the main theorem.) (2). Two metrics $d,d'$ on $S$ are inequivalent (meaning that $T_d\ne T_{d'}$) iff there exists $U\subset S$ which belongs to one of $T_d,T_{d'}$ but not the other. Without ...


2

I'm going to use the notation I'm familiar with, in which an $n$-simplex is denoted $[v_0,\dots,v_n]$, and the same $n$-simplex with $v_i$ removed is denoted $[v_0,\dots,\hat{v_i},\dots,v_n]$. The boundary operator is $$\rho_n[v_0,\dots,v_n]=\sum_{i=0}^n (-1)^i[v_0,\dots,\hat{v_i},\dots,v_n].$$ You seem to claim that $\rho_2[0,2]=[0]+[2]$ because ...


0

@Rolf Hoyer quote: You want to show that any arbitrary open set in one topology will be open in the other. I don't understand how that is sufficient. For example, I know that Euclidean metric and my "Clock" metric (see below) do not generate the same topology, on the set $[1,13)$. However the open ball $B_{2}^{clock}(1)=(11,13) \cup [1,3)$, is open ...


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 ...


1

You can understand $W_n$ as a CW-complex, so you have one $0$-cell $e^0_1$ and $n$ copies of $1$-cells $e^1_i$ ($i=1,\dots, n$) and any $m$-cell with $m>1$, so it means that if you want the fundamental group, you have $n$ diferent generators (one for each $1$-cell) and you don't have any restriction, so it means that you obtain that the fundamental group ...


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 ...


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 ...



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