# Tag Info

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You can't recover a sheaf from its stalks alone. For instance, all vector bundles $\mathcal{E}$ (of fixed rank $r$) on a variety $X$ have the same stalk at any given point $x \in X$: $$\mathcal{E}_{X, x} \cong \mathcal{O}_{X, x}^{\oplus r}$$ Stalks provide local information, whereas a sheaf encompasses not only local information but also information about ...

4

Related to QiaochuYuan's comment. It might seem a little tricky but maybe you can use something out of it. If you do not want to go through orbifold maybe this discussion could help you : $\pi_1$ and $H_1$ of Symmetric Product of surfaces Instead of looking topologically $\mathbb{T}^n/S_n$ you could say it is an orbifold (i.e. almost everywhere a manifold ...

4

I really like this proof because it uses a pretty neat trick: Declare an equivalence relation on $B$ by saying $a \sim b$ if there exists a bijection between $p^{-1}(a)$ and $p^{-1}(b)$. Try to prove that the equivalence class $[x]$ of $x \in B$ is both open and closed. Since $B$ is path-connected (and hence connected), this implies that $[x] = B$, which ...

3

I like the idea of using an algebraic model of a covering map, and one which has long been available is that of a covering morphism $q: H \to G$ of groupoids, dating back independently to C. Ehresmann and P.A. Smith, see also P.J. Higgins (1964) and his Categories and Groupoids, and the book by P. Gabriel and M. Zisman. The condition for this is that for ...

3

The Mobius band cannot be extended to a closed surface. Added: To answer the additional question, suppose that $\Sigma \subset \mathbb{R}^3$ is the given orientable surface. Then there is an embedding $f : \Sigma \times [0,1] \to \mathbb{R}^3$ such that $\Sigma = f(\Sigma \times 0)$. So then $\Sigma$ extends to a closed surface $$f\biggl(\bigl(\Sigma ... 3 f : S^1 \to S^1 be a map without fixed points. Consider the homotopy$$H(s, t) = \frac{(1-t)f(s) - ts}{||(1-t)f(s)-ts||}$$between f and the anitpodal map -\text{id}, which is well-defined since f(x) \neq x for all x. Compose this with the homotopy$$H(s, t) = e^{i\pi (1-t)} s$$between -\text{id} and the identity map \text{id}. Thus, f \sim ... 3 In general it really depends on how well you understand f and its action on the cohomology of X. Here are two nice special cases where you only need to understand the action of f on a single cohomology group and then everything else is determined by the cup product. Tori Let \Gamma be a lattice in \mathbb{R}^n. (You can take \Gamma = ... 3 Assume p is a quotient map: As any open set U \subset Y is open in Y \iff p^{-1}(U) is open in X, we have that p is continuous. For any saturated open subset, V \subset X, consider p(V). As V is saturated, we can write$$V = \cup_{y \in p(V)}p^{-1}(y)$$That is:$$V = p^{-1}(p(V))$$Thus as p^{-1}(p(V)) is open, p(V) must be too, and ... 3 First of all notice that 1-dimensional manifolds generically embedded in n-manifolds are unlinked for n\geq4. As you already noticed your space A := S^5 \setminus (S_1 \cup S_2 \cup S_3) is homeomorphic to \mathbb{R}^5 \setminus (\mathbb{R} \cup S_2 \cup S_3). This space is homotopically equivalent to (\mathbb{R}^5 \setminus \mathbb{R}) \vee ... 3 The use of Tietze transformations to prove that \langle a,b;(ab)^2\rangle\cong \mathbb{Z}*\mathbb{Z}_2 is correct. There are no mistakes in that. The mistake is to claim that \langle a,b;(ab)^2\rangle is a presentation of the projective plane. In fact, the relation \langle a,b;(ab)^2\rangle\cong \mathbb{Z}*\mathbb{Z}_2 together with the Classification ... 3 One can define characteristic classes in a more general context. For any topological group G, there is a universal principal G-bundle (henceforth just G-bundle) EG \to BG. BG is called the classifying space of G, because isomorphism classes of principle G-bundles over X correspond bijectively to homotopy classes of maps f: X \to BG, the ... 2 Let us denote the orbit space obtained from the S_n-action on X \times X \times \cdots \times X as SP^n(X). Given a continuous map f : X \to Y, SP^n induces a map SP^n(f) : SP^n(X) \to SP^n(Y). In particular, SP^n : \mathbf{Top} \to \mathbf{Top} is a functor. It can be proved that SP^n takes homotopic maps to homotopic maps, which means if ... 2 The nice concept to use here is that of coequaliser. Thus the diagram$$\coprod_{(i,j)\in I\times I} U_i\cap U_j \rightrightarrows^a_b\coprod_{i\in I} U_i \to^c X is a coequaliser in the sense that a map $f: X \to Y$ is completely determined by maps $f_i: U_i \to Y, i \in I$ such that $f_ia=f_ib$ for all $i$. The condition of an open cover is ...

2

The first relation: take any vector space with subspaces $C, Z\supseteq B$. The natural surjection $Z/B \to (Z+C)/(B+C)$ has kernel $(Z\cap C +B)/B$ which is isomorphic to $Z\cap C /B\cap C$ so the isomorphism you want follows from the first isomorphism theorem. $\partial^0$ is the map formed as the direct sum of the maps $C_{dp}/C_{d,p-1}\to ... 2 The Mayer-Vietoris sequence goes$\cdots \rightarrow H^{n - 1}(U) \oplus H^{n - 1}(V) \rightarrow H^{n-1}(U \cap V) \rightarrow H^{n}(M) \rightarrow H^{n}(U) \oplus H^{n}(V) \rightarrow 0$. Note that since$V \cong \mathbb{R}^{n}$we have$H^{n}(V) = 0$. We also already know that$H^{n-1}(U \cap V)$and$H^{n}(M)$are both$\mathbb{R}$, so in particular they ... 2 You more or less have the idea. Since every point of$X \subseteq Y$is collapsed to a point, and in$Y$every point of$S^{n-1} = \partial D^n$is mapped to$X$, the quotient$Y/X$is precisely$D^n / S^{n-1}$, which is$S^n$. More precisely, you can prove that$Y/X = (X \sqcup D^n) / {\approx}$where$t \approx x$for all$t \in S^{n-1}$and all$x \in ...

1

This is true in complete generality by just checking the universal property of the colimit (which is in your case also called a coequalizer, as already mentioned by Ronnie Brown): A cocone $f$ over the diagram of your question gives a family $\{f_i\colon U_i\rightarrow Y\}_{i\in I}$ of continuous maps, such that that the restrictions $f_i|_{U_i\cap ... 1 A handy way to see this is to consider the long exact sequence$...\to\pi_1(G)\to\pi_1(G/H)\to\pi_0(H)\to\pi_0(G)\to...$you find$\pi_1(G)=0$and$\pi_0(G)=0$because$G$is simply connected and$\pi_0(H)=H$because$H$is discrete. This just give$\pi_1(G/H)=H$. 1$\newcommand\Z{\mathbb{Z}}$Let$P = \mathbb{CP}^2 \times \mathbb{CP}^2$be the cartesian product and$S = \mathbb{CP}^2 \vee \mathbb{CP}^2$be the wedge sum. Then the smash product is$\mathbb{CP}^2 \wedge \mathbb{CP}^2 = P / S$. Since$S \subset P$is a sub-CW-complex,$\tilde{H}^n(P/S) \cong H^n(P,S)$. It is well known that$H^*(\mathbb{CP}^2) = ...

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You are right about the first part. If $p\omega$ is nullhomotopic in $A$, then the homotopy to the constant loop lifts to a homotopy between $\omega$ and the constant loop. This is because of the lifting properties of covering maps. Note that a covering map $p:\tilde X\to X$ restricts to a covering map $p^{-1}(A)\to A$ for every subset $A$ of $X$. Regarding ...

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For convenience, we may assume that the maps $\phi_i$ are inclusions of closed subspaces $X_i\subseteq Y_i$. These maps induce a closed inclusion $\phi:X_1\times X_2\times I\hookrightarrow Y_1\times Y_2\times I$. Seeing $X_1*X_2$ as a quotient of $X_1\times X_2\times I$ via the quotient map $q_X$ identifying $(x_1,x_2,0)\sim(x_1,x'_2,0)$ as well as ...

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Start the ordinary construction of the mobius band: gluing a rectangle $R$ with left side of $R$ identified to right side of $R$ (with a flip). Next, cut the rectangle into pieces, lets say $R_1,R_2,R_3$. Next, reconstruct the Mobius band by doing all the identifications at one time: the right side of $R_1$ is glued to the left side of $R_2$ (with no ...

1

With field coefficients you have that cohomology is the dual of homology. By linear algebra, we get an isomorphism. Proof for my statement is the fact that $Ext_{R}(H_i(M;R),R)=0$ for a field $R$ (because dualizing is exact for vector spaces), which you apply to the universal coefficient theorem. About the reasoning in the first part, you were completely ...

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