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13

In fact there are no submersions from an even sphere $S^{2n}$ into any smooth manifold $M$ of strictly smaller dimension. The pullback of the tangent bundle of $M$ along such a submersion would be a nontrivial quotient bundle of the tangent bundle of $S^{2n}$, but the tangent bundle of $S^{2n}$ has no nontrivial quotient bundles. This is because its Euler ...


10

I can prove there aren't any submersions $S^4 \to S^2$ or $S^5 \to S^2$. First, we want to apply Ehresmann's theorem. Because $S^m$ is compact any submersion $f: S^m \to S^n$ for is a proper submersion; we need to show that it's surjective. If $f: S^m \to S^n$ was not surjective, you could puncture $S^n$ at one of the points not in the image of $f$ to get a ...


7

This is a classical problem in algebraic topology, the Steenrod problem, which was more or less solved by Thom. Thom showed that this is true $\bmod 2$, rationally, and integrally if $k \le 6$. Integrally for $k \ge 7$ there are counterexamples; the obstructions involve Steenrod operations for odd primes. (It's harder to glue a bunch of simplices into a ...


7

The problem here is that the two maps "$p^{-1}$" for $U_{i-1}$ and $U_i$ may not agree on all of $U_{i-1}\cap U_i$, although they do agree at $F(y_0,t_i)$. For example, for the covering space $p:{\mathbb R}\to S^1$, if we use a cover of $S^1$ by two open intervals $U_\alpha$ and $U_\beta$ then it is impossible to choose lifts $p^{-1}:U_\alpha \to \tilde ...


5

The disc is a red herring here. You can write down a lot of surjective maps $D^2 \to S^1$; one might be $(r, \theta) \mapsto e^{2\pi i r}$. This is not a fluke. Indeed, it's indicative of a much wider phenomenon. Let $X$ be a subspace of $\Bbb R^n$. Then $X$ is a continuous image of $[0,1]$ if and only if $X$ is compact, connected, and locally connected. ...


5

Every $S^1$ bundle can be linearized: it's the unit sphere bundle of a 2-dimensional real line bundle. This is because in the smooth category $\text{Diff}(S^1)$ deformation retracts onto $O(2)$, so you may choose your bundle's cocycles to be linear (and then just define a vector bundle with the same cocycles). It's harder to show, but still true, that ...


4

The key to answering this question is the clutching construction. Let $p_1, p_2$ be antipodal points on $S^2$ and define $U_i = S^2\setminus\{p_i\}$. Note that $U_i$ is homeomorphic to $\mathbb{R}^2$ (via stereographic projection) and therefore contractible. Suppose $E$ is an $S^1$-bundle over $S^2$, then as $U_1$, $U_2$ are contractible, $E|_{U_1}$ and ...


4

Yes. Actually there is a more general result. If $X=\bigcup_{i=1}^n A_i$ where each subspace $A_i$ is contractible, then the product of any $n$ elements in $H^*(X)$ vanishes. Since $\Sigma X$ is a union of two cones, each of which is contractible, the result follows. To establish the general result, simply note that $H^*(X)\cong H^*(X, A_i)$ as $A_i$ is ...


4

If $X \to Y$ is a finite Galois cover with Galois group $G$ then it's known that the induced map $H^{\bullet}(Y, \mathbb{Q}) \to H^{\bullet}(X, \mathbb{Q})$ is injective and induces an isomorphism $$H^{\bullet}(Y, \mathbb{Q}) \cong H^{\bullet}(X, \mathbb{Q})^G.$$ (This has nothing to do with $X$ and $Y$ being compact or manifolds.) So here the answer is ...


4

You're working too hard. Think about the cell decomposition of $\Bbb{RP}^n$; it's obtained from $\Bbb{RP}^{n-1}$ by adding an $n$-cell by the projection $S^{n-1} \to \Bbb{RP}^{n-1}$. We may pick the point you're deleting to be in the interior of the $n$-cell (otherwise, pick a homeomorphism of $\Bbb{RP}^n$ that takes it there; the homeomorphism group of a ...


4

Not in general. These subgroups will be conjugate but might not be equal. You can check this by taking a path between your points in $\widetilde{X}$ and projecting it down to a loop in $X$. It is true if $p$ is a regular covering. Also, you don't need the "locally path connected" assumption for any of these arguments.


4

Spanier's book is relatively old (so I know it does not quite answer your question), but excellent. It uses category theory from the get-go. Riehl's "Categorical homotopy theory" is very well-written, though it may be a bit too advanced if you hadn't seen a bit of algebraic topology already. Riehl's book is focused on the categorical aspect via Quillen model ...


4

I suggest Peter May's Concise course on algebraic topology. You will find e.g. categorical formulations (and proofs) of the van Kampen theorem and the classification of covering spaces.


4

The closest thing I've found is Strom's Modern Classical Homotopy Theory, although I haven't read much of it. Chapter 1 is called Categories and Functors, so that's a good start. This is the only introductory algebraic topology textbook I know of that explicitly uses the language of homotopy limits and colimits.


3

If $n$ is an integer, the complex power map $z \mapsto z^{n}$ extends to a map of the Riemann sphere $\mathbf{C} \cup \{\infty\}$ having degree $n$. (It's understood that if $n = 0$, then $z^{n} = 0$ for all $z$, while if $n > 0$ then $\infty^{n} = \infty$, and if $n < 0$, then $\infty^{n} = 0$ and $0^{n} = \infty$. These conventions do not constitute ...


3

Here is a solution which doesn't use any machinery from algebraic topology. Let $X'$ denote the countable set of points removed from $S^2$. Let $x \in X'$ and let $X = X'\setminus\{x\}$. By stereographic projection from $x$, $S^2\setminus X'$ is homeomorphic to $\mathbb{R}^2\setminus Y$ where $Y$ is the image of $X$ under the stereographic projection. As ...


3

Let me write $B^n A$ for $K(A, n)$. The Yoneda lemma implies that $[B^n \mathbb{Z}, B^{np} \mathbb{Z}]$ is precisely the set of natural transformations $H^n(-, \mathbb{Z}) \to H^{np}(-, \mathbb{Z})$. An obvious candidate for such a natural transformation is the $p^{th}$ cup power $$H^n(-, \mathbb{Z}) \ni \alpha \mapsto \alpha^p \in H^{np}(-, \mathbb{Z})$$ ...


3

Let's try a small value of $n$ first to see how it works. The first nontrivial case is $n=2$, so take $c : [0,1]^2 \to A$ a singular $2$-cube. Its boundary is $$\partial c = \sum_{i=1}^2 \sum_{\alpha \in \{0,1\}} (-1)^{i+\alpha} c_{(i,\alpha)} = -c_{(1,0)} + c_{(1,1)} + c_{(2,0)} - c_{(2,1)}.$$ If $u : [0,1] \to A$ is a singular $1$-cube, its boundary is: ...


2

As defined in Section 2.1 of Hatcher's book, a $\Delta$-complex structure on a space $X$ is a collection of maps $\sigma_\alpha : \Delta^n \to X$, satisfying a list of properties. About a page after the beginning of that definition, Hatcher gives the notation $e^n_\alpha = \sigma_\alpha(\text{interior}(\Delta^n))$, and the terminology that $\sigma_\alpha$ ...


2

First, note that if $\min\{p, q\} = 0$, then by definition $C$ is empty and so has zero connected components. For other signatures, the count is given by the following: Proposition If $\min\{p, q\} > 1$, then $C$ has one connected component. If $\min\{p, q\} = 1$ but $\max\{p, q\} > 1$, then $C$ has two connected components. If $p = q = 1$, then ...


2

The map $F: E\times [0,1]\to E$ defined fiberwise by $(b,x)\mapsto (b,tx)$ is continuous and is a deformation retraction from the identity on $E$ (in $t=1$) to $B$ (in $t=0$), where $B$ is identified with the zero section of $E$. A deformation retraction induces cohomology isomorphism. (For the other direction, as the identity on $E$ and the projection to ...


2

Here is an elementary, easy, self-contained proof which, as a friend reminded me, results from a discussion we had a few months ago. I could kick myself for forgetting about it yesterday ... Consider two subschemes $X_1,X_2\subset \mathbb P_k^n=\operatorname {Proj}(S)\quad (S=k[T_0,\cdots,T_n]) $. From the exact sequence $$0\to S/I(X_1)\cap I(X_2)\to ...


2

There is a nice general method for calculating the arithmetic genus of a reducible curve like this where a bunch of curves are being glued transversely. It uses the (equivalent) description of the arithmetic genus of a curve $X$ over a field as the dimension of the cohomology group $H^1(\mathcal O_X)$. So suppose $C_1$ and $C_2$ are curves over a field. Let ...


2

The required arithmetic genus is $$p_a(X)=0$$ You can see it by using a formula published in 1957 by Hironaka in the article: On the arithmetic genera and the effective genera of algebraic curves (Theorem 2, page 190). The formula is $$ p_a(C)=\pi(C)+\sum_P\delta (P) -(r-1)$$ The sum is over the singular points $P\in C$ and $\delta (P)=\dim (\tilde ...


2

Yes, I think it's crucial for the proof that $I$ is a CW complex (or at least has good properties). Indeed, in the category of spaces, directed limits do not necessarily commute with product by a fixed space. And another way of saying that $X$ is coherent with its skeleton is to say that $X$ is the colimit $X = \operatorname{colim}_{n \ge 0} X^n$. So in ...


2

When $n>1$, $S^n$ is simply connected, and $H^1_{dR}(M) = 0$ for any simply connected manifold. (Hint: For any closed $1$-form $\omega$ and any closed curve $\gamma$, we have $\displaystyle\int_\gamma\omega = 0$.) Now, in order to deal with $\Bbb RP^n$, consider the $\Bbb Z/2\Bbb Z$ action given by the deck transformations. Show that any exact invariant ...


2

I presume you're asking how to equip a given space with a CW-structure. There is no general procedure to do it. One usually tries to express his space into smaller and simpler looking spaces, and then obtain a CW-structure on the whole space from "patching-up" the CW-structure on those smaller spaces. For example, Given a pair of CW-complexes $(X, A)$ with ...


2

Here is a slightly more general proposition: For any connected manifold $M$ of dimension $\geq 2$, and a countable subset $X$, $M - X$ is path connected. Let $p,q$ be two points in $M - X$. Let $\gamma$ be a path between them in $M$. By compactness, definition of a manifold, and a cardinality argument, $\gamma$ can be decomposed as a sequence of paths ...


2

One way to do this is to take the degree $n$ map $f : S^1 \to S^1$ given by $e^{iz} \mapsto e^{inz}$ and extending this to the map $g = f\times I : S^1 \times I \to S^1 \times I$ given by $(e^{iz}, t) = (e^{inz}, t)$, compose with an appropriate homeomorphism $S^1 \times I \cong S^2 - \{(0, 0, 1), (0, 0, -1)\}$ and finally extending continuously to the ...


2

The suspension of the map $z\to z^n$ of $\mathbb{S}^1$ is a map of $\mathbb{S}^2$ of degree $n$. Geometrically, this wraps each latitudinal line of the sphere around itself $n$ times and maps the north pole to itself and the south pole to itself. Analytically, this is the extension of the complex function $z\to z^n$ to the Riemann sphere $\mathbb{S}^2$. ...



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