# Tag Info

7

Take the group $\mathbb{Z}/3\mathbb{Z} = \{0,1,2\}$ under addition, with the topology consisting of $\emptyset$, $\{0,1,2\}$ and $\{1\}$. Now negation is not continuous, since the preimage of the open set $\{1\}$ is $\{2\}$ which is not open. Similarly, addition is not continuous, since the preimage of the open set $\{1\}$ is $\{(0,1),(1,0),(2,2)\}$ which ...

7

Let $b$ be the bottom point of the sphere, i.e. $(0,0...,0,1)$. For any $x$ in the ball let $f(x)$ be the midpoint of the line segment with endpoints $b$ and $x$.

5

I would presume that "constant point" refers to a solution to $f(x)=x$ - that is, what would typically be called a fixed point. The function $re^{iax}\mapsto re^{ibx}$ certainly does have a constant point at $0$, so it wouldn't work. A really simple example would be to consider that the operation of contraction towards a point (i.e. scaling) is continuous, ...

5

By the Universal coefficient theorem (for cohomology) and the assumption the $M$ is simply connected, we have $$H^2(M)=\operatorname{Hom}(H_2(M),\mathbb{Z})\oplus \operatorname{Ext}(H_1(M),\mathbb{Z})= \operatorname{Hom}(H_2(M),\mathbb{Z})$$ which is torsion free, hence free abelian (because it is finitely generated). By Poincaré duality, $H^2(M)\cong ... 4 For the first part, let$i\colon S^1\to D^2$be the inclusion of the circle into the unit disk and, since$f$is null-homotopic, let$\tilde{f}\colon D^2\to S^1$be an extension of$f$to the whole disk (which exists). Since$f$has no fixed points, and the image of$\tilde{f}$lies within$S^1$, what can we say about$i\circ \tilde{f}\colon D^2\to D^2$and ... 4 I think this answers your question, but I'm not entirely sure. Hopefully some bits of it are useful to you--let me know! I think, first and foremost, you have a bit of confusion regarding the two uses of the word 'differential' here. On one hand, when you speak of the differentials as$g=\dim H^0(X,\Omega^1_{X/\mathbb{C}})$you're speaking of algebraic ... 4 I can't think of any direct disadvantages other than the complication of the definition. But then, engineers, analysts and applied mathematicians aren't usually bothering with exponentiability of spaces anyway-although mathematical physicists might, in trying to formalize some of the "spaces" that arise in physics. 3 I think the main idea is that of a "category which is adequate and convenient for all purposes of topology" as I wrote in the Introduction of my first, 1963, paper, available from here. That is, the idea of looking at the properties of the category of the objects one is studying is a useful one. In analysis, one may need something different. See the book ... 3 Yes, this is true. First, since$\partial M \hookrightarrow M$is not$\pi_1$-injective, by the Loop Theorem there is a properly embedded 2-disc$(D,\partial D) \hookrightarrow (M,\partial M)$such that$\partial D$does not bound a disc in the torus$\partial M$. Second, letting$N$be a regular neighborhood of$\partial M \cup D$, the surface$\partial ...

3

Not all proper subsets of $S^n$ are contractible (unless $n = 0$ obviously). Take any two unequal points $x \neq y \in S^n$, then $\{x,y\} \subset S^n$ is not contractible. What is true is that $S^n \setminus \{ x \} \cong \mathbb{R}^n$ is contractible, so if the map isn't surjective, it factors through a contractible space so it's nullhomotopic.

3

Yes, just the same. You can easily check what the completion of $k(1/t)=k(t)$ is. Best to look first at the $(t)$-adic completion of the ring $k[t]$: this is the ring $k[[t]]$ of formal power series in $t$ over $k$. Then the associated complete ring is $k((t))$, the (finite-tailed) formal Laurent series over $k$. When $k$ is finite, your field is a member of ...

3

A necessary condition for an $n$-dimensional manifold $M$ to be immersible into $\mathbb{R}^{n+k}$ is that there must be a $k$-dimensional vector bundle $N$ on $M$ (the normal bundle) such that $T \oplus N$ is trivial, where $T$ denotes the tangent bundle. (By the Hirsch-Smale theorem, this necessary condition is actually sufficient when $k \ge 1$.) So we ...

3

Lemma: Show that if $A$ is a retract of $B^2$, then every continuous map $f : A \to A$ has a fixed point. Proof: Suppose that $A$ is a retract of $B^2$, then by definition there exists a continuous map $r : B^2 \to A$ such that $r(a) = a$ for all $a \in A$. Let $f : A \to A$ be an arbitrary continuous map. Define $g : B^2 \to B^2$ by $g = j \circ f \circ ... 3 The Jordan-Schoenflies theorem states that the inside and outside of a Jordan curve are homeomorphic to the inside and outside of a standard circle in$\mathbb{R}^2$. You can read more in this paper. 2 I'll use the more standard terminology "fixed point" rather than "constant point". Consider the universal covering map$p : \mathbb{R} \to S^1$,$p(t)=exp(2 \pi i t)$. Let$\tilde f : \mathbb{R} \to \mathbb{R}$be a lift of$f$. Let$d = \deg(f)$, and I'll consider three separate cases. In the first two cases I'll prove that$\tilde f$has a fixed ... 2 Sometimes, you actually have to look at the topology. A chain$c$in$H_1(X, A)$is a collection of edges (with coefficients, but those will turn out to be irrelevant); the boundary is a collection of pairs-of-points in$A$. Since$A$is path connected, for each such pair we can find a path in$A$that connects them. The sum of all these paths, with the ... 2 Expanding on the comment of Adeel, you have to exploit the following easy fact about orthogonal classes of arrows in categories: Let$F\dashv G$be two adjoint functors between categories$\mathcal{C}\leftrightarrows\mathcal{D}$; then$Ff\perp g$in the category$\mathcal D$(i.e.,$Ff$has the LLP with respect to$g$in$\cal D$) if and only if$f\perp Gg$... 1 The easiest way, for me at least, is first understand the dual of each abelian group in the complex, then construct the dual to the boundary operator. Let$G$be an abelian group. The dual to$G$is just the group whose elements are homomorphisms$G\to\mathbb{Z}$. Note that if$G$is freely generated by the set$\{g_i|i\in I\}$, then a homomorphism ... 1 In the first definition you have for what it means for$U$to be evenly covered, the mutually disjoint open sets are your sheets.$(1)$and$(2)$say the same thing, except for$(1)$, you would probably add "a sheet over$U$via$p$" (or "by$p$", as you said) to emphasize that the homeomorphism is given by$p$restricted to$V$. 1 Set$Y=[0,1]$. Use path lifting to define$f$. Then apply the homotopy lifting theorem. 1 You just need to apply the homothopy lifting thorem twice. Since$p(e_0)=F(0,0)$, by lifting the map$F_0=F \restriction_{\{0\} \times [0,1]}$you get a map$G_0: \{0\} \times [0,1] \to C$such that$G_0(0,t)=F(0,t)$and$G(0,0)=e_0$. Now you can lift the whole map$F$to a map$G:[0,1] \times [0,1] \to C$such that$(p \circ G)(s,t)=F(s,t)$and ... 1 One possible reason is that if we allowed infinite formal sums then it would be impossible to define the boundary map. Suppose our simplicial complex consists of all simplices of the form$[v_0,v_i]$with$i\geq 1$. Consider the formal sum $$\sum_{i=1}^{\infty}{[v_0,v_i]}$$ The boundary of this linear combination would be ... 1 First, let me make the weaker claim that$\mathbb{HP}^{\infty}$is not naturally an H-space. The natural H-space structures on$\mathbb{RP}^{\infty}$resp.$\mathbb{CP}^{\infty}$come from the fact that they classify isomorphism classes of real resp. complex line bundles, which naturally have group structures given by taking the tensor product. This no ... 1 The fundamental group and higher homotopy groups are defined only for a space with a chosen base point, a structure which looks like a space, but is a bit more. I explain in this presentation, given in Paris last June, how the problem with the van Kampen theorem and the fundamental group of the circle led me to the use of groupoids and then higher ... 1 Let$X$be a subspace of$\mathbb R^n$and let$Y$be some topological space. Consider the inclusion function$j : X \to \mathbb R^n$where$x \mapsto x$. Suppose you have$\varphi : (X, x_0) \to (Y, y_0)$a continuous map that extends to$\mathbb R^n$, that is, there exists$k : (\mathbb R^n, x_0) \to (Y, y_0)$such that$k \vert_X = \varphi$. But recall ... 1 Le$f:P^n\to P^n$be the map you describe and let$f^*:H^*(P^n)\to H^*(P^n)$be the induced map on cohomology. Since$f^*$is a map of rings and$H^*(P^n)$is generated as a ring by its degree two component, it is enough to describe$f^2:H^2(P^n)\to H^2(P^n)$. Moreover,$H^2(P^n)$is a free$\mathbb Z$-module or rank$1$, generated by one class$\alpha\in ...

1

There's another definition of boundary for maniforld, which are spaces where every point has a neighborhood homeomorphic to $\mathbb R^n$ or $\mathbb R_+^n$, for a fixed $n$. The boundary of a manifold is the set of points with neighborhood homeomorphic to $\mathbb R_+^n$, and indeed for $\mathbb R^n$, there is no boundary.

1

Usually, topologists talk about boundaries of a subset of another set. So, if your $\mathbb{R}^4$ sits in a larger space -- for example, in $S^4$ -- it may have a boundary (one point, for example). There is another definition of boundary for manifolds: in this sense, $\mathbb{R}^4$ has no boundary. As far as I know, physicists have various theories and ...

1

Every set has a (possibly empty) boundary, whether or not it is open. Let $X$ be a topological space, say $\Bbb R^4$ with standard topology. If $A$ is an arbitrary subset of $X$, then the boundary of $A$ is the set of points that are "near" to both $A$ and $X \setminus A$ and is sometimes defined by $\partial A = \operatorname{cl}(A) \setminus ... 1 Prove that for any$x_0 \in X$, the set$\{ x \in X : |p^{-1}(x)| = |p^{-1}(x_0)| \}$is clopen in$X$. It follows that if$X$is connected, this is the whole set. Now it's a problem of definitions: the number of sheets of$p$is defined to be the cardinality$\kappa$such that$|p^{-1}(x)| = \kappa$for every$x\$. It doesn't always exist: sometimes the ...

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