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9

Removing a subcomplex of codimension 3 or higher from a manifold does not change the fundamental group. Therefore, in this example the fundamental group is $Z^6$. Edit: Let $M$ be an $n$-manifold and $K\subset M$ be submanifold of codimension $\ge 3$. (I will do this in smooth category but it works in topological setting too: It suffices to assume that $K$ ...


5

Read about both simultaneously in the same book: Differential forms in Algebraic Topology ! (Bott was one of the best twentieth century geometers, and it shows in this extraordinary book)


4

There is a canonical homomorphism from $\hom_S(A,R) \otimes_R \hom_S(B,R)$ to $\hom_S(A \otimes_S B,R)$. The question is if this is an isomorphism of $R$-modules. It is true when $A$ and $B$ are finitely generated and projective (the standard proof works - show that $S$ works and that the class of "good" modules is closed under direct sums as well as direct ...


3

Every closed 4-manifold is homeomorphic to a CW complex if and only if every compact 4-manifold with nonempty boundary is homeomorphic to a CW complex.


3

Of course this is not true as stated because there exist nonorientable closed $3$-manifolds, for example $S^1 \times \mathbb{RP}^2$. Once you have orientability $w_1 = v_1 = 0$.


3

$K(G \rtimes H, 1)$ fits into a fibration sequence $$K(G, 1) \to K(G \rtimes H, 1) \to K(H, 1).$$ So for example one can access the homology and cohomology using the Serre spectral sequence. See this answer for some context.


3

The quotient map $q:X\times [0,1]\to CX$ which sends $X\times\{0\}$ to a single point restricts to a quotient map $q':X×(0,1]\to CX-\{P\}$ since $X×(0,1]$ is open and saturated in $X×[0,1]$. Since $q'$ is injective, it is a homeomorphism. Now a homotopy $(CX-P)×I\to CX-P$ is determined by a homotopy $X×(0,1]×I→X×(0,1]$. In particular, a retracting homotopy ...


3

If $Y$ is aspherical and path connected, then a map $f : X^{(2)} \to Y$ can be extended to $\tilde{f} : X \to Y$. This is a special case of the extension lemma (lemma 4.7) in Hatcher's Algebraic Topology: Given a CW pair $(X, A)$ and a map $f : A \to Y$ with $Y$ path-connected, then $f$ can be extended to a map $X \to Y$ if $\pi_{n−1}(Y) = 0$ for all $n$ ...


3

The trick is to embed the category of abstract simplicial complexes inside the category of symmetric simplicial sets (= functor $\mathbf{F}^\mathrm{op} \to \mathbf{Set}$, where $\mathbf{F}$ is the category of positive finite cardinals): this can be done by sending an abstract simplicial complex $X$ to the symmetric simplicial set ...


2

Your statement is only true for normal covering spaces $\hat X\to X$, i.e. $\pi_1(\hat X) \subset \pi_1(X)$ is normal. Then we get the isomorphism $Deck(\hat X) \cong \pi_1(X) / \pi_1(\hat X)$, which is very interesting, have a look at its geometric interpretation. Furthermore there are threefold coverings which have trivial deck transformation group. Of ...


2

The two are fairly closely linked, and it wouldn't be a bad idea to study them concurrently. That said, I think if you have to pick one, go with algebraic topology first. The first reason is historical. Before the modern theory of differential topology was developed, algebraic (and combinatorial) methods were employed. It is usually much easier to calculate ...


2

If we think of a map $F \in Hom(V \otimes V,k)$ as a matrix with $A_{ij} = F(e_i \otimes e_j)$, where $\{e_j\}$ is a basis of $V$, then in the finite dimensional case, we can recover any map $\hom(V \otimes V, k)$ as a finite sum of matricies of the form $vw^T$, which is what an element of $\hom(V, k) \otimes \hom(V,k)$ essentially is under the canonical ...


2

Let $p:\mathbb{R}^6\to T^3\times T^3$ be the universal covering map, which I consider as reducing each coordinate mod $1$. Consider the pre-image $p^{-1}(\Delta)\subset\mathbb{R}^6$. This is the set of $(t_1,\ldots,t_6)\in\mathbb{R}^6$ such that $(t_4,t_5,t_6)-(t_1,t_2,t_3)\in\mathbb{Z}^3$. In other words, $$ p^{-1}(\Delta) \cong ...


2

I assume by a "closed equivalence relation" you mean one whose equivalence classes are all closed sets. The equivalence classes are $A$ and all the individual points of $X-A$. So if $A$ is closed and $X$ is Hausdorff then yes, the equivalence relation is closed. But the only way the equivalence relation could be open would be if $X-A$ is a discrete subset of ...


1

In terms of coordinates $\pmb{x} = (x_1, ..., x_n) \in (\mathbb{R}/ \mathbb{Z})^n$ for $T^n$, consider the map $f: T^n \times (T^n \setminus \lbrace \pmb{0} \rbrace) \rightarrow (T^n \times T^n) \setminus \Delta$ given by $f(\pmb{x}, \pmb{y}) = (\pmb{x}, \pmb{x}+\pmb{y})$. The map $f$ is continuous with continuous inverse $f^{-1}(\pmb{x}, \pmb{y}) = ...


1

First, a general element of $S\otimes G$ is of the form $\sum \alpha_i\otimes g_i$. But that is not the problem. The boundary of this is $\sum (\partial \alpha_i)\otimes g_i$ and you are assuming that its a cycle so it is zero in $S\otimes G$, however this does not imply that it is zero in $B\otimes G$. Here is an example of this phenomena, which arises ...


1

The space $H^1(K) = \mathbb{Z}^2$ is generated by the Poincaré duals $\alpha = A^*$ and $\beta = B^*$ to $A$ and $B$, respectively. (I'm working over $\mathbb{Z}_2$ throughout, so that $H^*(K)$ is actually $H^*(K, \mathbb{Z}_2)$. For reasons of dimension, the only products you need to compute in the ring $H^*(K)$ are $\alpha^2, \alpha \beta$, and $\beta^2$. ...


1

For those who want to compute it all out, if we use the delta-complex below instead of the one I gave in the question we have: Let $\phi \in C^0(K,\mathbb{Z}_2)$ be dual to $v\in C_0(K,\mathbb{Z}_2)$, $\alpha, \beta, \gamma \in C^1(K,\mathbb{Z}_2)$ be the dual elements to $a,b,c\in C_1(K,\mathbb{Z}_2)$, respectively, and $\mu,\lambda\in ...


1

I can suggest two nice books covering basics of the above subjects. Hopefully they give you some perspective. Topology and Geometry, by Breadon is book primarily on algebraic topology but it treats the subject using differential manifolds, so you probably enjoy reading its 2-3 chapters. Here is the link. Foundations of Differentiable Manifolds and Lie ...


1

So clearly SW classes are not trivial for non-orientable manifolds as mentioned in the previous comment. But if you are interested in some geometry you can try proving(!) or find a proof of the fact that orientable 3 manifolds are parallelizable, which would then give an easy proof of vanishing of SW classes for orientable manifolds.


1

As suggested by @Lee Mosher, using the Mayer-Vietoris sequence with $A = \phi(S^1 \times \mathbb{D}^2) \cup \{ small \textrm{ } nbhood \}$ and $B = S^3-A$, we get that $A \cup B = S^3$ and $A \cap B \simeq T^2$. If we want to use Mayer-Vietoris to solve for just the first homology group say, then the sequence $$ H_2(A \cup B) \to H_1(A \cap B) \to H_1(A) ...


1

A bit late on this, but this type of question is considered by Chadwick and Mandell in http://arxiv.org/abs/1310.3336, although they give conditions on when the map is $E_2$. This suggests that in general the problem is difficult!


1

The wikipedia page "Fundamental polygon", specifically the subsection entitled "group generators", has a serious mathematical error. You cannot derive a presentation for the fundamental group from the fundamental polygon using the side labels in the manner described on that page (and which you have copied), unless all of the vertices of the polygon are ...


1

If $X$ is a subvariety of $\mathbb{C}^N$ of complex codimension $\leq 2$, then the fact that $\mathbb{C}^N\smallsetminus X$ is simply connected follows from the normality of $\mathbb{C}^N$ and the Zariski-Nagata purity theorem; see SGA2 X Theorem 3.4(i). You can get both $\pi_1$ and $\pi_2$ trivial by using general position. See Lemma 2.4 of this paper ...


1

You went in the wrong direction. Use that $H_k(X^n,\emptyset) \cong H_k(X^{n+1},\emptyset) \cong H_k(X^{n+2},\emptyset) \cong \cdots$, together with the usual limiting argument based on the fact that every chain is supported in some finite dimensional skeleton.


1

I'm no expert on the matter and this was intended as a comment but it ended up to be too long. I think what is meant in the last sentence of the first paragraph you cited is that $h_0h_{\infty}^{-1}$ induces a clutching function $f$ for $E'$ by restricting to $E\times S^1$. The last sentence in the second paragraph seems a bit strange to me... As far as I ...


1

A $1$-form is just a map that assigns to each point in your manifold a linear functional on the tangent space at that point, denoted $T_pM$. That is, for $p\in M$ we have that $\omega(p):=\omega_p\in T_p^\ast M$. For simplicity, if you'd like just replace $M=\mathbb{R}^n$ so that $T^\ast_pM=T_pM=\mathbb{R}^n$ and so a $1$-form just eats $n$-tuples and ...


1

You can define a "simplex" as having an orientation, thus getting an easier answer. A $k$-simplex is the convex hull of a set of $k+1$ points. But what does it mean for a set to have $k+1$ points? That there is a bijection from $\{1,2,\dots,k+1\}$. So simply define "$k$-simplex" in terms of a map $\{1,2,\dots,k+1\}\to\mathbb R^n$ and you can pick an ...



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