# Tag Info

0

I'd like to add a more general approach to this problem. Given a subcomplex $L$ of a simplicial complex $K$, the homomorphism $C_n(L)\to C_n(K)$ which sends any simplex in $L$ (as a generator of the simplicial chain group) to the same simplex in $K$ gives a chain map which fits into a commutative diagram below the relative chain group $C_n(K,L)$ is ...

2

To elaborate on my comment, applying $\text{Ext}^{\bullet}(A, -)$ to the short exact sequence $0 \to \mathbb{Z} \to \mathbb{R} \to S^1 \to 0$ produces the long exact sequence $$0 \to \text{Hom}(A, \mathbb{Z}) \to \text{Hom}(A, \mathbb{R}) \to \text{Hom}(A, S^1) \to \text{Ext}^1(A, \mathbb{Z}) \to 0$$ where we can ignore the rest of the sequence because ...

1

Define a map $f:C_n(K)\to\mathbb{Z}$ by sending $\sigma$ to $1$ and every other $n$-simplex to $0$. Since $\sigma$ is not a face of any $(n+1)$-simplex, $f$ vanishes on any boundary, and thus induces a map $C_n(K)/B_n(K)\to\mathbb{Z}$. Restricting this to the subgroup $H_n(K)\subseteq C_n(K)/B_n(K)$, we get a map $\psi:H_n(K)\to \mathbb{Z}$. You might ...

2

The third property (with $G=\mathbb{Z}$) tells you that if $C$ is a finite cyclic group, then $\operatorname{Ext}(C,\mathbb{Z})\cong C$. Furthermore, any finitely generated torsion abelian group is a direct sum of cyclic groups. So $T$ is a direct sum of cyclic groups, so by the first and third properties, $\operatorname{Ext}(T,\mathbb{Z})\cong T$. Note, ...

1

Your confusion stems from the fact that you are using the wrong definition of $Z^n$ and $B^n$. They are not defined to be $\ker(\partial^n)$ and $\operatorname{im}(\partial^{n-1})$. Rather, they are defined to be the duals of $Z_n$ and $B_n$; that is, $Z^n=\operatorname{Hom}(Z_n,G)$ and $B^n=\operatorname{Hom}(B_n,G)$. The map $i^n$ is then the map that ...

1

Let me write $\varphi$ for your map taking $[g]$ to $[\tilde{g}]$. There are three things you have to check to show $\varphi$ is a bijection: $\varphi$ is well-defined: if $[g]=[g']$, then $[\tilde{g}]=[\tilde{g}']$. $\varphi$ is injective: if $[\tilde{g}]=[\tilde{g}']$, then $[g]=[g']$. $\varphi$ is surjective: if $[f]\in[\mathbb{S}^1:X]$, then there ...

0

Hint: This space is contractible, can you see the contraction? What does this say for the fundamental group?

3

observe that $\mathbb{R^2}$ is the universal cover of torus, and since $f^*$ is the zero map, so by map lifting lemma, you can lift $f$ in $\mathbb{R^2}$, and since $\mathbb{R^2}$ is contractible, so image is contractibe , i.e image is homotopic to zero. Now compose the homotopy with covering map will give a null homotopic map in $\mathbb{T^2}$.

5

It is not true in general that a map that induces 0 on the level of fundamental groups is null-homotopic. (Take the identity map of $S^2$!) You need further argument, such as: Because $f_*$ is zero, you can factor $f$ through the universal cover $\Bbb R^2 \to T^2$. Because $\Bbb R^2$ is contractible, the map is null-homotopic. E: The same argument works ...

1

assume $x_0$ is the identity element...let $f$ and $g$ be two loops at $x_0$ then $f*e_{x_0} \approx e_{x_0}*f$ where $e_{x_0}$ define constant loop. Now we can write $f*g =(f*e_{x_0}).(e_{x_0} *g) \approx (e_{x_0}*f).(g*e_{x_0}) = g*f$

0

This is also proved in Topology and Groupoids (as it was in the 1968 edition, "Elements of Modern Topology"); this has some pictures of the crucial mapping cylinder construction $M(f) \cup X$ which, if $i: A \to X$ is a cofibration, is a useful model of the adjunction space $B \cup _f X$ for $f: A \to B$. Here is a coloured picture of the homotopy as Fig ...

0

I got the solution to this problem. Every space is locally simply connected because of the definition of covering. Also, we can consider $U \subset X$ simply connected (and thus a trivializing neighbourhood). Then , the preimage of $U$ is: $$\varpi^{-1}(U) = \sqcup_{i} V_i$$ Where every $V_i$ is homeomorphic to $U$. In the same way is easy to see that: ...

2

If you've thought about the logarithm in the context of complex variables, you know it must be defined with branch cuts or else be a multivalued function. If you want to "graph" the multivalued function, what you have is a Riemann surface. Indeed, Riemann surface theory is a natural outgrowth of complex analysis when you want to see the topology behind ...

0

This is not a complete answer, but... A non-trivial fundamental group space means (roughly speaking) a more complicated space. Since the covering space share many properties with the original space, and because "good" spaces (with some reasonables hypothesis) always have universal covering (a covering with trivial fundamental group), to pass to the ...

1

The CW-complex being constructed in your notes is not a triangle (and your approach to constructing a triangle as a CW-complex is perfectly correct). Rather, it is a triangle with the three edges of the triangle all glued together, with the orientation indicated by the arrows in the picture. There is only one vertex, because when you glue the edges like ...

1

I'm not sure I follow the part of your question asking whether the attachment should result in $(k-1)$-spheres. When we attach a $k$-cell (via its boundary) to a point it's really just the same as identifying the boundary $S^{k-1}$ of $D^k$ to a point, so we end up attaching a copy of $D^k/S^{k-1}$ at a point. Then $D^k/S^{k-1}=S^k$, so we are indeed ...

1

For $1)$ call the space $X$. Then $S^2$ intersects $X$ at three points. If we denote the set of intersection points $P$, then $S^2\setminus P$ is a deformation retract of $X$. By stereographic projection, we know $S^2 - \{pt\}$ is homeomorphic to $\mathbb{R}^2$ so $S^2\setminus P$ is homeomorphic to $\mathbb{R}^2$ with two points removed. This space ...

1

Write your polynomial $p(z) = \prod_{i=1}^n (z - a_i)$, where the $a_i$ are the roots ("zeroes") of the polynomial $p$. Then define the homotopy $H(z,t) = \prod_{i=1}^n (z - t a_i)$; basically it "sends all the roots to zero". This is clearly continuous, and $H(z,0) = z^n$ while $H(z,1) = p(z)$, so $p$ and $z \mapsto z^n$ are homotopic.

2

Let $\varepsilon >0$ and let $x\in \mathbb{R}^N$, then for any $y\in B_{\varepsilon}(x)$ you have a homeomorphism $$\tau_y^{x,\varepsilon} \colon B_{\varepsilon} (x) \to B_{\varepsilon} (x)$$ which maps $x$ to $y$ and extends to the identity on the boundary, hence is can be extended to $\mathbb{R}^N$ and we will call this map $T_y^{x,\varepsilon}$ Let ...

0

Firts assume that $X$ is simply connected. Then $X'= X\times D$, $D$ a space with the discrete topology. Therefore $X"= X\times D'\times D$ for another discrete space $D'$, and the result follows. The general case follows from the hypothesis : let $x\in X$ and $U$ a simply connected neigbourhood of $X$, the previous argument applied to $U\subset X$ ...

0

I'll first give a brief excursus about the different ways to look at a delta complex A general Delta complex $X$ can be described combinatorially as a sequence of sets $X_0,X_1,\dots$ together with maps $d_i:X_n\to X_{n-1}$ for any $i\in[n]=\{0<\dots<n\}$ such that $$d_j d_i = d_i d_{j+1}, \qquad \text{whenever }\; i\le j$$ whenever $j\ge i$. ...

0

A lot of confusion arises in this problem due to an index shift performed during the calculation. I'm going to change the notation used slightly, but I'll hopefully define everything below (let me know if I miss something). First, I'll write the i-th face map $d^i:\Delta^{n-1}\rightarrow\Delta^n$: ...

6

I see this as an additive/multiplicative notation issue, exactly as you say. It's like when one talks about exact sequences of abelian groups or modules, one often writes $$0 \to A \to B \to C \to 0$$ whereas one may more likely write $$1 \to X \to Y \to Z \to 1$$ for an exact sequence in the category of all groups.

4

From the comment: $(1)$ Find a continuous map $f : \mathbb S^1 \times \mathbb S^1 \to \mathbb S^1$ so that $f_*$ is surjective. $(2)$ Let $\gamma : \mathbb S^1 \to \mathbb{RP}^2$ represent the generator of $\pi_1(\mathbb{RP}^2)$. Then $\gamma_*$ is surjective. $(3)$ The composition $\gamma \circ f$ is one of the example as $(\gamma\circ f)_* = \gamma_* ... 0 If$A$is a CW-complex and$\varphi:S^{k-1}\to A$is a cellular map, then the space$A\cup_\varphi e^k$obtained by attaching a map via$\varphi$is a CW-complex. The condition that$\varphi$is cellular means that its image is contained in the$(k-1)$-skeleton$A^{k-1}$. So actually, when building up$A\cup_{\varphi} e^k$inductively as a cell complex, ... 3 I'll prove the statement I have in my comment. That any CW-complex with the homology of$S^n$suspends to a space homotopy equivalent to$S^{n+1}$for$n\geq 1$. If$X$is such a CW-complex, by Mayer-Vietoris,$SX$has the homology of$S^{n+1}$. As$X$is$(0)$-connected,$\pi_1(SX)=0$by the Freudenthal suspension theorem. By applying Hurewicz repeatedly, ... 1 We may define i: I → I to be the identity map, i¯: I → I to be i¯(s) = 1-s and e: I → I to be e(s) = o for all s∈I. Then we may define H: IxI → I by H(s,t) = (1-t)(i∗i¯)(s)+te(s) for all (s,t)∈IxI. Since I is a convex subset of real numbers, H is well defined. Then you may check f。H is a path homotopy between f∗f¯ and ex0. 2 More generally, if$F:Ab\to Ab$is any functor which preserves addition of maps, then it sends split exact sequences to split exact sequences. This follows from the following theorem: Theorem Let$A\stackrel{i}{\to} B\stackrel{p}{\to} C$be a pair of maps of abelian groups. Then the following are equivalent: There exist maps$q:B\to A$and ... 5 There is the long exact sequence in homology of the pair$(X, X \setminus p)$(everything is with$\mathbb{Z}$-coefficients unless indicated otherwise): $$0 \to H_n(X \setminus p) \to H_n(X) \to H_n(X, X \setminus p) \to H_{n-1}(X \setminus p) \to H_{n-1}(X) \to 0$$ (recall that the local homology groups satisfy, by excision,$H_n(X, X \setminus p) = ...

2

$H_0(X,x_0)$ is the $0$-homology of the chain complex $$\frac{C_1(X)}{C_1(x_0)} \to \frac{C_0(X)}{C_0(x_0)} \to 0$$ Compare: $\tilde{H}(X,x_0)$ is the $0$-homology of the "extended" chain complex $$\frac{C_1(X)}{C_1(x_0)} \to \frac{C_0(X)}{C_0(x_0)} \to \frac{\mathbb{Z}}{\mathbb{Z}}\simeq 0.$$

1

Let $y = f(x)$. Then $f$ determines an isomorphism from $\pi_1(X, x)$ to $\pi_1(X, y)$. If we had $y = x$ then this would be an automorphism of $G$, but we don't. So what can we do instead? We can fix a path between $x$ to $y$, which also induces an isomorphism from $\pi_1(X, x)$ to $\pi_1(X, y)$, and use this path to turn the isomorphism induced by $f$ ...

2

If you move the ends of the segments around inside the body, you can show that the space is homotopy equivalent to $T^2\vee S^1\vee S^1$. If you wanted to make this rigorous, just find a contractible subspace $A$ of the torus which contains the end points of each segment (a tree in the torus with the ends of the line segments at its leaves will do) and then ...

2

Since it preserves orientation, your homeomorphism is going to be a rotation of the simplex. Take a narrow annulus $A$ surrounding $\Delta$ with one boundary component $\partial_0A$ equal to the boundary of the simplex $\Delta$. Now we want to extend $g$ from $\partial_0A$ to $A$ in such a way that it restricts to the identity on $\partial _1A$. To do this ...

3

Every odd-dimensional manifold has vanishing Euler characteristic, so $$0 = \chi(M) = b_0 - b_1 + b_2 - b_3.$$We have $b_0 = 1$, and $b_3 = 0$ since $M$ is nonorientable. Hence, $b_1 > 0$ and thus $H_1(M, \mathbb{Z})$ is infinite.

2

[This argument is stolen from the end of this answer (which handles the case $q=1$).] Let $1\leq i\leq n-1$ and $\alpha\in H_i(M,\mathbb{Z})$. By Poincare duality, there exists $\beta\in H^{n-i}(M,\mathbb{Z})$ such that $\alpha=z\cap\beta$. Since $H^{n-i}(S^n,\mathbb{Z})=0$, $f^*(\beta)=0$. Thus $0=f_*(i_n\cap ... 2 The transfer homomorphism of the orientation cover, composed with the pushforward, is multiplication by$2$. The image is free abelian since the$(n-1)$-integral homology of the orientation cover is by my answer here, so the only torsion in$H_{n-1}(X, \mathbb{Z})$is$2$-torsion. Applying the universal coefficient theorem for$H_n$with$\mathbb{Z}/2$... 0 The first$1$-skeleton has$5$vertices and$10$edges. Using the canonical$CW$structure of$S^2$together with the fact that the Euler characteristic is invariant under$CW$decomposition we have, $$5 - 10 + F =2$$ $$F =7$$ Each face has at least$3$edges so we count at least$3F$edges. Every edge is counted twice (it is on the border of two ... 2 Yes, this is true. The natural map$H_n(X,\mathbb{Z})\otimes \mathbb{Z}_p\to H_n(X,\mathbb{Z}_p)$is an isomorphism for$X=S^n$and$X=M$. We thus have the following commutative diagram, where the vertical maps are isomorphisms:$$\require{AMScd} \begin{CD} H_n(S^n,\mathbb{Z})\otimes\mathbb{Z}_p @>{f_*\otimes 1}>> H_n(M,\mathbb{Z})\otimes ... 2 You should be able to find a proof in most books covering homology theory. A specific reference is Theorem 4.59 of Hatcher's Algebraic Topology (p. 399-402). As Mike Miller commented, the idea of the proof is to show that any (co)homology theory can be computed by a "cellular chain complex" associated to that theory, and that these cellular chain complexes ... 1$S$is a topological surface with Euler Characteristic$\chi(S) = 2$. Now$G$acts on$S$, then$S/G$is covered by$S$and so$\chi(S) $is divisible by$\chi(S/G)$. Thus$|G| = 1$or$2$. Remark That$G$is a subgroup of the group of deck transformations implies that$G$acts properly discontinuously on$S$: For all$s\in S$,$s\in M$and so there is an ... 3 Suppose you have an oriented 3-manifold with an embedded$\Bbb{RP}^2$. The normal bundle can't be trivial lest the whole 3-manifold be non-orientable; so taking a tubular neighborhood you have a 3-manifold with$S^2$boundary (the total space of the unit disc bundle of the tautological bundle); call this$M_1$. By the irreducibility hypothesis the boundary ... 1 HINT: Suppose that$\{W_1,\ldots,W_n\}$is a finite open cover of$X$such that$W_k\cap Y$is closed in$W_k$for$k=1,\ldots,n$. If$Y$is not closed in$X$, there is some point$x\in(\operatorname{cl}_XY)\setminus Y$. Choose$k$such that$x\in W_k$.$Y\cap W_k$is closed in$W_k$, and$x\notin Y\cap W_k$, so there is a relatively open$U\subseteq W_k$... 1 If$Y$is closed, then every$W_\alpha \cap Y$is also closed, by definition of subspace topology. If$Y$is not closed, then we have$Y' = \bar Y \setminus Y \neq \varnothing$, where$\bar Y$is the closure of$Y$. This$Y'$must intersect$W_\beta$for some$\beta$(since the$W_\alpha$cover$X$). We then have that the closure of$W_\beta \cap Y$in ... 1 First just consider the torus with one segment attached. Pick a nice loop going through the segment. Now pick a regular neighborhood of that loop as one open set and the complement of the loop as the other set. Boom, you got yourself two sets satisfying Van Kampen's theorem. Observing that the intersection is topologically trivial you get immediately a free ... 1 A closed orientable$n$-manifold is called homology sphere if it has homology (or equivalently cohomology) of a sphere. Note that for a closed connected orientable 3-manifold we have always$H_0M\cong \mathbb Z \cong H_3M$and also$H_2M \cong H^1M \cong Hom(H_1M,\mathbb Z)$. You see that$H_1M=0$forces it to be a homology sphere. 5 Yes. First, recall that a suspension has no nontrivial cup products. By Poincaré duality, it follows that the integral cohomology of$M$is torsion in degrees$1$through$n-1$, and that the same is true of the integral homology. From universal coefficients we know that$H_{n-1}$is torsion-free, which here implies that it's trivial. By Poincaré duality ... 3 Since$M$is orientable, without boundary and connected, we have$H_0(M;\mathbb{Z}) = H_3(M;\mathbb{Z}) = \mathbb{Z}$. Using Poincaré duality, you know that the torsion subgroup of$H_2(M;\mathbb{Z})$is isomorphic to the torsion subgroup of$H_{3-2-1}(M;\mathbb{Z}) = H_0(M;\mathbb{Z}) = \mathbb{Z}$and thus is zero. Using Poincaré duality again, you know ... 3 Consider the long exact sequence of the pair$(M,\partial M)$and notice that Lefschetz duality gives an isomorphism$H_k(M)\cong H^{n-k}(M,\partial M)\cong H_{n-k}(M,\partial M)$, using that$M$is orientable and the coefficients are over a field. Consider the truncated exact sequence$H_m(\partial M)\to H_m(M)\to H_m(M,\partial M) \to H_{m-1}(\partial ...

3

Tensor product is right exact: if $0 \to V' \to V \to V'' \to 0$ is an exact sequence of abelian groups, then $M \otimes V' \to M \otimes V \to M \otimes V'' \to 0$ is an exact sequence too (I'm looking at tensor product over $\mathbb{Z}$; if $f : X \to Y$ is a morphism, then the induced morphism $M \otimes X \to M \otimes Y$ is given on generators by $m ... 0 This only answers the index 2 case (and does not use covering spaces). Since every subgroup of index$2$is normal this can be obtained by calculating how many surjective homorphisms there are from$F_2$to$\mathbb{Z}_2$. Since there are only four maps from a two element set$\{x,y\}$to$\mathbb{Z}_2$there are exactly four homorphisms from$F_2\$ to ...

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