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5

Here $|\alpha|$ refers to which dimension cohomology group $\alpha$ is in, i.e. $|\alpha|=1$ means that $\alpha\in H^1(\mathbb{R}P^n;\mathbb{Z}_2)$. Note that the asserted isomorphism is supposed to be not just an isomorphism of rings but an isomorphism of graded rings, so you have to specify a grading on the ring $\mathbb{Z}_2[\alpha]/(\alpha^{n+1})$. ...


4

This is most probably the pullback of the covering map $\mathbb{R} \to S^1$. In a more general context, suppose that $p : E \to B$ is some covering space, and $f : X \to B$ is any continuous map. Then the linked Q&A shows that $$f^*E := E \times_B X = \{ (e,x) \in E \times X \mid p(e) = f(x) \}$$ is a covering space via the projection $q : f^* E \to X$, ...


4

For a simple counterexample, take $X=S^1$ and $Z$ to be a point. Then $\chi(X\setminus Z)=1\neq \chi(X)-\chi(Z)=-1$. Morally, what's going wrong here is that you should consider $\chi(X\setminus Z)$ to be $-1$ rather than $1$, since it is made up of only a single (open) $1$-cell and no other cells. But if you take the standard homological definition of ...


3

The whole Section 1.2 of Hatcher's book is dedicated to van Kampen's theorem, where nonabelian fundamental groups are abundant. The more-or-less canonical example of the wedge sum $S^1 \vee S^1$ is in particular described, and I find the explanation in the introduction of the section rather clear. You have two loops, $a$ and $b$, and it's not possible to ...


3

Suppose that you have a homotopy equivalence $f : X \to Y$ with homotopy inverse $g : Y \to X$, where $Y$ is some discrete space (of any cardinality, finite or infinite). The space $X$ is compact, thus $f(X) \subset Y$ is compact; but the only compact subspaces of a discrete space are finite, so $f(X)$ is finite. Since $f$ and $g$ are inverse homotopy ...


2

If the composition $$f\circ \phi:\mathbb{S}^n\to X_n\to Y$$ is nulhomotopic, then it extends to the disk, i.e., there exists $\bar f:D^{n+1}\to Y$ such that if $i:\mathbb{S}^n\to D^{n+1}$ denotes the inclusion, the obvious diagram commutes: $f\phi = \bar fi$. Now recall that $X_{n+1}=X_n \cup_\phi D^{n+1}$ is the pushout of the diagram \begin{array} ...


2

It sounds like the question arises from two points of confusion. 1) If you think of * as an element of $X_0$, then you need to remember that an element of $X_0$ gives rise, via the degeneracy maps, to an element in each $X_n$. (Play around with the simplicial identities to see why you get a well-defined element in $X_n$, regardless of which degeneracy maps ...


2

Personally, the way I think of this is to look at the following decomposition of $\mathbb{C}^2$ into two sets, and see how they fit together into the quotient. Let $U = \{(x, y) \in \mathbb{C}^2 \mid y \neq 0\}$ and let $P = \{(x, 0) \in \mathbb{C}^2\}$. Note that these are disjoint and their union is all of $\mathbb{C}^2$. When we quotient $P$ by the ...


2

With regard to intuition, imagine tying together two broom handles A,B by a rope going once round A and then once round B; this is not the same as, i.e. cannot be deformed into, going once round B and then round A. With regard to proof, I prefer to use the van Kampen theorem for the fundamental groupoid $\pi_1(X,S)$ on a set $S$ of base points found in ...


2

The answer is yes for the following reason: $\omega_n$ is the Euler class mod 2 Now use e.g. Theorem 4.7 here which says $e(\nu_N)([N])$ counts number of intersections, or you argue that the Thom class of $\nu_N$ in $M$ is the Poincaré dual of $N$ (follows from this exercise). Hence, by pulling back to the cohomology of $N$ the result follows. You ...


2

The first Chern class of the tautological bundle over $P(E_p)$ is a generator of $H^2(P(E_p))$. Lets call this $a=-i^{*}(x)$ with $i:P(E_p)\rightarrow P(E)$ the inclusion of the fiber. The cohomology ring of $P(E_p)$ is $H^*(P(E_p))=Z[a]/a^{n}$ (note that $P(E_p)\cong \mathbb{C}P^{n-1}$). So an additive basis of the cohomology of the fiber is ...


2

Let us assume $X$ and $Y$ are Hausdorff (otherwise, I'm not sure what "the one point compactification" is supposed to mean). Then $X\wedge Y$ is compact Hausdorff, being the quotient of the compact Hausdorff space $X\times Y$ by a closed equivalence relation (the equivalence relation is closed because it is the union of the diagonal in $(X\times Y)^2$ and ...


1

The Solution is correct but more complicated than necessary. One can just show that $X$ and $\partial D$ are not homotopy equivalent by Computing their fundamental group or First homology. Namely, $\partial D$ is homeomorphic to $S^1$, so $\pi_1=H_1=Z$. But $X=T^2-D^2$ is homotopy equivalent to the wedge of 2 circles, so $\pi_1=Z*Z$ and $H_1=Z\oplus Z$.


1

How does the uct arise? Because we have the natural epimorphism $H^k(M;G) \twoheadrightarrow Hom(H_k(M),G)$. UCT tells you that this splits, hence for $G=\mathbb Z$, as the kernel is torsion, you obtain for $k=3$ that this epimorphism is non-trivial. But as $H_3(M;\mathbb Z)$ is a finitely generated abelian group, $0\neq Hom(H_3(M),\mathbb Z) \to Hom(H_3, ...


1

Yes. The condition on homotopy groups (vanishing in higher degree, isomorphism in lower degree) is satisfied for the products. And the product of two fibrations is a fibration, so all involved maps are fibrations.


1

The inclusion $* \times X \subset S^1 \times X$ is a cofibration (the singleton $*$ is a sub-CW-complex of $S^1$, hence $* \subset S^1$ is a cofibration, and by this question the product of a cofibration with an identity map is a cofibration), thus: $$H^*(S^1 \times X, * \times X) \cong \tilde{H}^*\bigl( (S^1 \times X) / (* \times X) \bigr)$$ (this is a ...


1

In order to avoid a long discussion, let me clarify this: the first proof is incomplete. Let $\alpha$ is a path from $f(x_0)$ to $g(x_0)$, then your $H$ is a homotopy from $f$ to a constant map $x\mapsto g(x_0)$. On the other hand, $g\circ F$ is another homotopy from $g$ to the constant map $x\mapsto g(x_0)$. Then you use the fact that being homotopic is an ...


1

"I mean for anyone who trusts the eye theorem is obvious." How can you trust your eye when you are unable to fully visualize general continuous curve, e.g. a fractal? If a point $z_0$ lies on a fractal that is a Jordan curve, then there is no way to tell in what direction the "inside" is. For this reason we must "be blind" and resort in our proof only to ...


1

There is in fact a number concerned with the minimal amount of open subsets to cover $M$ which satisfy contractability in $M$ --- the Lusternik Schnirelmann category $cat(M)$. This gives you (at least with my definitiong of a chart) $$ cat(M)\leq \eta(M). $$ There are a lot of interesting techniques presented in the literature for this, which might be ...


1

You're right that $\iota_{\mathbb{F}_2,0}$ is not the multiplicative unit. However, it still is its own square. Indeed, for any space $X$, every element of $H^0(X,\mathbb{F}_2)$ is its own square. Even more strongly, this holds on the chain level: any singular $0$-cochain on a space with coefficients in $\mathbb{F}_2$ is its own square. Indeed, a ...


1

It seems to me that you have all the correct ideas about topology. Topology works best as a mixture of intuition and precision. The fuzzy intuitive dreaming is fun, but it needs to interact correctly with the precise and rigid part. For instance, dreamy intuition can be a helpful step in formulating a precise mathematical description of what is true and what ...


1

It might be easier to think of a topological group as a topological space with a group structure rather than a group with a topology. Let $X$ be a topological space. This means that certain subsets of $X$ are in a family $\tau\subset\mathcal{P}(X)$ such that the usual axioms apply. Now assuming the axiom of choice, there is a group structure on $X$. That ...


1

For your second paragraph, it might be helpful to see an example of a topological space with a discontinuous group operation. The topological space is $X = \mathbb{R}$. I will pick a discontinuous bijection of $\mathbb{R}$: $f(x) = x$ if $x \ne 0,1$, $f(0)=1$, $f(1)=0$. Using this bijection, I will define a group operation $\oplus$ on $X$, having identity ...


1

Perhaps try thinking about matrices. You know from linear algebra that the set of invertible $n \times n$ matrices is a group with group operation matrix multiplication. If you write out the functions that express the entries of the product of two matrices in terms of the entries of the matrices you can see that they are continuous (since they involve just ...


1

For a homotopist's proof see the comment I made above. There is a simpler proof however. Take a component $X$ of $\Omega S^1$. It corresponds to some integer $n$ because $\pi_0(\Omega S^1)\cong \Bbb{Z}$. That means that $X$ is the subspace of loops going $n$ times around the circle. By the general theory of universal covers, you can prove that $X$ is ...


1

To elaborate on Qiaochu's comment: Suppose $H:A \times I \to A$ is a homotopy from the identity to the constant map on $x_2$. Then $\lambda=H(x_1,-)$ is a path from $x_1$ to $x_2$. I claim that $P_{0 \to x_1, 1 \to x_2}(A)$ is homotopy equivalent to $\Omega A$, the based loops at $x_1$, which is contractible. Consider the maps $$f:P_{0 \to x_1, 1 \to ...


1

If $E=\{1,2\}$, $B=\{3\}$ and $p(x)=3$ for both $x\in E$, then $p^*\colon H^0(B) \to H^0(E)\colon\mathbb{Z}\to\mathbb{Z}^2$ is the map $n\mapsto (n,n)$ which is not surjective.



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