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7

0) The excellent mathematician you evoke has as family name (=surname) tom Dieck and as first name Tammo: tom is part of his surname and has nothing to do with Tom, the endearing form of Thomas. 1) Your idea of "finding a book that starts with a formal treatment of the basics of category theory and moves to more advanced/specialized concepts in a ...


5

The space $S^1\vee S^1\vee S^2$ is the wedge sum of two circles and one sphere. In particular this space is also known as mouse space. It is quite easy to see that the fundamental group of $S^1\vee S^1\vee S^2$ is $\mathbb{Z}*\mathbb{Z}\cong\langle a,b| \emptyset \rangle$ ( you can use the Seifert-VanKampen theorem to see it ); whereas the fundamental ...


4

Yes, being simply-connected is preserved under homeomorphisms. Suppose $f : X \to Y$ is a homeomorphism and $X$ is simply connected. First, path-connectedness is a topological invariant, so $Y$ is path-connected too. Now, let $\gamma : S^1 \to Y$ be any (continuous) loop. Then $f^{-1} \circ \gamma : S^1 \to X$ is a continuous loop, and $X$ is ...


3

Pick a basepoint $x_0 \in U\cap V$. Let $p:I \to X$ be some loop starting at $x_0$. Then we can write $p$ as a finite product of paths $p = f_1 \cdot \ldots \cdot f_n$, where each $f_i$ is entirely contained in either $U$ or $V$. WLOG we can assume that each $f_{2i}$ is entirely contained in $U$ and each $f_{2i+1}$ is entirely contained in $V$ (so they ...


3

Because to obtain our chain complex for a space $X$ we look at arbitrary maps $\Delta^n \to X$ which are not necessarily nice, i.e. singular. In contrast we look at nice embeddings when dealing with simplicial homology.


3

You run into two technical problems with this attempt to describe homology, but they're not "obvious" problems. (In fact, your description is at least partially what motivated Poincare to define homology in the first place.) The first is the restriction to submanifolds. There are too many technical details about embedding submanifolds: how many submanifolds ...


3

Yes, and in fact you do not need $X$ to be an H-space: the coproduct $\psi$ of $H_*(X)$ exists anyway (if the ring of coefficients is a field), and the definition of a group-like element does not involve the product. So take $X$ to be any space, and define $S \subset H_*(X)$ as in your question (the set of classes in $H_*(X)$ satisfying $\psi(a) = a \otimes ...


3

Hodge theory is a vast subject, especially since Deligne and Griffiths revolutionized it and algebraized it around 1970 with the introduction of variations of Hodge structures and mixed Hodge structures, axiomatic approaches which found amazing applications in algebraic geometry. At the most elementary level I would say that Hodge theory is a refinement ...


3

The point is that if $p:X\rightarrow Y$ and $q:Y\rightarrow Z$ are covering maps in this sense, but some $z\in Z$ has infinitely many $y_i\in Y$ with $q(y_i)=z$, then each $y_i$ may have some neighborhood $U_i\subseteq Y$ "evenly covered" by $p$ (I take this to mean the inverse image of $U_i$ is a union of parts each mapped isomorphically to $U_i$) but as ...


2

This is a supplement to Tyler's answer (it should be a comment, but it's too long). As Tyler mentions, the Hauptvertmutung (any two triangulations have a common refinement) is false for topological spaces with dimension greater than 2. This breaks the sentence "simplicial (co)homology is an approximation of oriented cobordism (co)homology theory in all ...


2

Sorry, this isn't a full answer, rather some algebraic hints at how you might relate the cross product to your intuition about the cup product (wrt intersections) and to linear algebra. It should really be a comment but it's too long. First, a quick review! The cross product, $\times$, is actually part of the cup product: $$x \smile y := \Delta^*(x \times ...


2

Given a continuous map $f : X \to Y$, there is an induced chain map $f_\# : C_n(X) \to C_n(Y)$ of singular chains. Since $C_n(X)$ is a free abelian group whose basis is the set of singular simplices $\sigma : \Delta^n \to X$, in order to define $f_\#$ all you need to do is to define it's value $f_\#(\sigma)$ on cach basis element, and then $f_\#$ is uniquely ...


2

It sounds as if you're confused about the difference between singular/simplicial homology and cellular homology. If you look at the homology chapter of Hatcher's book again, you'll see that these subjects are covered in some detail. The relevant details are as follows: The homology of a space is a deep property of that space that is hidden behind a ...


1

In Hatcher's Algebraic Topology, these are called characteristic maps, and the restriction $\varphi\colon \partial\Delta^n\to \Delta$ to the boundary is referred to as the attaching map. In my experience, this usage is fairly standard.


1

I expect the restriction to open $n$-simplices is just to remove any potential ambiguity from the fact that closed $n$-simplices may overlap, while their interiors do not.


1

Take $A, B$ subspaces of $X$ whose interiors cover $X$. There are three induced homomorphisms we need to understand:$$H_n(A\cap B)\xrightarrow{\quad\Phi\quad}H_n(A)\oplus H_n(B)\xrightarrow{\quad\Psi\quad}H_n(X)\xrightarrow{\quad\partial\quad}H_{n-1}(A\cap B)$$ The map $\Phi$ comes from the inclusions of subspaces $\iota_A:A\cap B \hookrightarrow A$ and ...


1

Firstly, why are you writing $\hom_{F-modules}(H_n(X;F),F)$? UCT implies $$H^n(X;F)=\hom_{F}(H_n(X;F),F)$$. Now, every $F$-module is a $F$-vector space and dual of a (finite dimensional) vector space is isomorphic to itself. So, $H^n(X;F)=\hom_F(H_n(X;F),F)=H_n(X;F)$, whenever $H_n(X;F)$ has finite rank. For the infinite rank case see this SE post.


1

Here is a solution, actually not involving homotopy groups at all: In the real projective space, we want to follow the orientation given in a base point, going through the nontrivial closed loop until coming back to the base point. To do so, we take the loop in $S^n$ which is the half-circle between two antipodal points; since $S^n$ is orientable, what ...


1

We will draw on the computation in Chapter 23, Section 3 of May's A Concise Course in Algebraic Topology that$$w(\mathbb{R}P^q) = \sum_{0 \leq i \leq q} \binom{q+1}{i} \alpha^i,$$where $$H^*(\mathbb{R}P^q ; \mathbb{Z}_2) \approx \mathbb{Z}_2[\alpha]/\alpha^{q+1}.$$Suppose that $q = 2^k - 1$. Then note that$$\binom{2^k}{i} \equiv 2 \text{ }(\text{mod }2)$$ ...


1

[Note: After the comments to this answer and the question, there have been major changes. Thanks to Najib Idrissi, the situation is much clearer now.] There are only slight differences in the definitions and strictly speaking, Hatcher's is more general. In addition to (equivalents of) Hatcher's axioms, May wants a Hopf algebra to be flat (a minor technical ...


1

The questioner is right to identify one of the inspirations for homology as coming from integration theory; this is emphasised in the article by S. Lefschetz in the book "History of Topology", edited I.M. James. It seems also that early articles on Betti nubers and torsion coefficients wanted to take "cycles modulo boundaries" but were not so clear about the ...


1

The comment I made above misunderstood the problem, and I'm sorry for that... To see they are the same, notice that $C_0(A)$ is not empty so the mapping $C_0(A)\stackrel{\epsilon}{\rightarrow}\mathbb{Z}$ is already surjective. Therefore $C_0(X,A)\stackrel{\epsilon}{\rightarrow}\mathbb{Z}$ is a zero mapping and its kernel is the same as $\ker \partial_0$.


1

Your claim will work fine. But you need to change the thing you're tensoring over back into a subalgebra! $F[\pi_0 X]$, the sub-$F$-algebra generated by $\pi_0(X)$, is the right choice. It might be easiest to justify by just showing that for $S$ a multiplicative subset of a commutative $F$-algebra $A$ and $G$ the free group on the monoid $S$, ...


1

The issue is that $a^*+b^*$ is not the generator of $H^1$, in fact it is not a cohomology class: $$(a^*+b^*)(dU) = (a^*+b^*)(a+b-c) = 2.$$ Instead, take $b^*+c^*$ as the generator, which is a cohomology class: $$(b^*+c^*)(dU) = (b^*+c^*)(a+b-c) = 0,$$ $$(b^*+c^*)(dV) = (b^*+c^*)(a+c-b) = 0.$$ Also, I believe that $\mathbb{Z}[x,y]/(x^3, 2x^2, xy, y^2)$ is not ...


1

I am not sure If I understand the definition correctly, but if I do, then take a look at the maps $$ f : (0,1) \rightarrow [0,1], x \mapsto x$$ and $$ g : [0,1] \to (0,1), x \mapsto \frac{1}{3} x + \frac{1}{3}.$$ These are embeddings and composing these we get $$ f \circ g : [0,1] \rightarrow [0,1], x \mapsto \frac{1}{3} x + \frac{1}{3}$$ and $$ g \circ f ...


1

There is a homology theory that looks like this called (co)bordism, and it is very interesting. It comes in many flavors depending on what kind of extra structure you ask for on the manifolds. The basic problem with your proposal is functoriality: the image of a submanifold need not be a submanifold. The correct definition of bordism fixes this by allowing ...



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