# Tag Info

7

No. Consider the case of the unit circle $S_1=\mathbb R/\mathbb Z$. The map $f(x)=x$ winds around the circle once, where $g(x)=x+x$ winds around the circle twice. They have the same image and the same endpoints, but are not homotopic. A somewhat more general example is to take any curve not homotopic to the trivial curve. Then, construct a new curve which ...

7

Sure, why not? If $X$ is an uncountable discrete set and $Y$ is any space, the projection $X \times Y \to Y$ is a covering map. If you don't like this, then you could instead construct a CW complex (which always have universal covers) $X$ with $\pi_1(X)$ uncountable. Then the universal covering map $\tilde X \to X$ has uncountably many slices. It just so ...

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.

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, $H_n(X, X \setminus p) = ... 5 It is not true in general that a map that induces 0 on the level of fundamental groups is null-homotopic. (Take the identity map of$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$is contractible, the map is null-homotopic. E: The same argument works ... 4 From the comment:$(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_* ...

4

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

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, ...

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 $H_m(\partial M)\to H_m(M)\to H_m(M,\partial M) \to H_{m-1}(\partial ... 3 You can take the trivial covering space$E = B \times F$where$F$is an uncountable set with the discrete topology with$\pi \colon B \times F \rightarrow B$the projection. Then$B$is evenly covered by the disjoint union$\bigcup_{f \in F} B \times \{ f \}$of open subsets of$E$homeomorphic to$B$. 3 No. Consider the paths $$f: [0, 1] \to S^1 : t \mapsto (\cos 2 \pi t, \sin 2 \pi t)\\ g: [0, 1] \to S^1 : t \mapsto (\cos 4 \pi t, \sin 4 \pi t).$$ These are both surjective onto the unit circle$S^1$, and have the same starting and ending points, but are not homotopic as paths in$S^1$. (They are homotopic as maps into$\mathbb R^2$, though!) 3 Any manifold book worth its salt should prove this. Put a Riemannian metric on$X$; then for every point in the boundary, there is a unique tangent vector in$T_p X$that is orthogonal to the boundary, of norm 1, and points inwards. This provides a trivialization of the normal bundle of$\partial X$in$X$. Now, how would one prove the tubular neighborhood ... 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$\mathbb{T^2}$. 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 The wedge sum is a quotient space of the disjoint union. That is, we take the disjoint union$X=\bigsqcup_{\alpha\in A} I_\alpha$, define an equivalence relation$\sim$on$X$by$x\sim y$iff either$x=y$or there exist$\alpha,\beta\in A$such that$x=1_\alpha$and$y=1_\beta$, and define the wedge sum to be the quotient space of$X$by this equivalence ... 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 ...

3

The rational cohomology $H^{\bullet}(G, \mathbb{Q})$ of a compact connected Lie group is the exterior algebra on some odd generators, the product of which lives in top cohomology. The number of generators $r$ is the rank. The map $g \mapsto g^{-1}$ acts by $-1$ on each generator, and so it acts on top cohomology by $(-1)^r$. Hence $g \mapsto g^{-1}$ reverses ...

2

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 $b_0 = 0$ since $M$ is nonorientable. Hence, $b_1 > 0$ and thus $H_1(M, \mathbb{Z})$ is infinite.

2

The problem is that you are performing an operation which is "not allowed": The cup product $\alpha\smile\alpha$ is a cohomology class in $H^2(X;\mathbb Z)$. Now such a class is determined only up to coboundaries: If $φ$ and $\psi$ represent the same class, then $φ-\psi$ is a coboundary, and that means it vanishes on cycles. So if $C$ is a cycle in $C_2(X)$, ...

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 $0=f_*(i_n\cap ... 2 By the universal coefficient theorem applied in degree$n$to$H_n(X, R) \cong R$, any$\mathbb{Z}$-tor of$H_{n-1}(X, \mathbb{Z})$vanishes, hence it is a flat, equivalently projective, equivalently free,$\mathbb{Z}$-module. 2 This is just the commutativity of the restriction maps with the cup product on the level of cochains combined with the fact that$\tilde{H}_\bullet(X) = H_\bullet(X, *)$. 2 Suppose you have an oriented 3-manifold with an embedded$\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$. By the irreducibility hypothesis the boundary ... 2 From the commutative diagram here,$H_\bullet(X, A \cup B) = H_\bullet(X, X) = 0$, so if we can show the left arrow is surjective, we are done. However, on each factor, it is just a restriction map, so this is clearly the case. 2 More generally, if$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$and ... 2 You should be able to find a proof in most books covering homology theory. A specific reference is Theorem 4.59 of Hatcher's Algebraic Topology (p. 399-402). As Mike Miller commented, the idea of the proof is to show that any (co)homology theory can be computed by a "cellular chain complex" associated to that theory, and that these cellular chain complexes ... 2 If you've thought about the logarithm in the context of complex variables, you know it must be defined with branch cuts or else be a multivalued function. If you want to "graph" the multivalued function, what you have is a Riemann surface. Indeed, Riemann surface theory is a natural outgrowth of complex analysis when you want to see the topology behind ... 2$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 $$\frac{C_1(X)}{C_1(x_0)} \to \frac{C_0(X)}{C_0(x_0)} \to \frac{\mathbb{Z}}{\mathbb{Z}}\simeq 0.$$ 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 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, ...

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