# Tag Info

1

Although you've got the answer by yourself, I would like to write an answer solving the problem with cellular homology, so that someone who asks the same question can find an answer here. I solved this problem a few month ago in an Algebraic Topology course as an exercise. Proof: Let $X = S^1\times S^1/ \sim$ be the space with the identifications: ...

0

If we were to imagine cutting the surface of both shapes along the "edges" (where in the Penrose triangle we ignore the triangular bends), then the resulting surface is one continuous piece in both cases, but there will be a different number of half-twists (three for the triangle, one for the torus). So in that regard, the two surfaces are not equivalent.

0

These are topologically both the same. Indeed both are homeomorphic to a solid torus $S^1 \times D^2$, and hence are homotopy equivalent to a circle: you can imagine shrinking each cross-section down to a point. The edges/faces thing is irrelevant from the point of view of topology. For example the unit square $[0, 1]\times [0, 1]$ is homeomorphic to the ...

2

You need only the following three facts: $H^1(X\sqcup Y) = H^1(X) \oplus H^1(Y)$, $H^1 (X\times \mathbb R) = H^1(X)$, and $H^1(\mathbb S^1) = \mathbb R$. The first and the third facts can be proved directly, while the second one (sometimes referred as the Poincare lemma) might be a bit harder to show. Everything are well discussed in the book ...

6

A non-orientable manifold has a connected double cover called the orientation double cover (see proposition 3.25 in Hatcher). Because it's a double cover, it induces an index-two subgroup of $\pi_1(M)$ (the subgroup of loops whose lifts are also loops). Quotienting out by this subgroup gives a nonzero homomorphism $\pi_1(M) \to \Bbb Z/2$, and hence (by the ...

0

If $X$ is truly associative (not just up to homotopy), then since the AW map is associative too, the composite will be associative. Otherwise, choose $x,y,z \in X$ such that $(xy)z \neq x(yz)$. If $u \in X$ is some point, let $\sigma(u) : \Delta^0 \to X$ be the associated zero-simplex (constant map equal to $u$). Then the image of $(\sigma(x), \sigma(y))$ ...

0

This doesn't really have much to do with equivariant cohomology. Consider (open) pathconnected subspaces $U,V$ such that $U\cap V$ is path-connected. Then $H^{0}(U)\cong H^{0}(V)\cong H^0(U\cap V)\cong \mathbb{R}$. These are generated by constant functions on $U,V$ and $U\cap V$. The map $\phi$ then maps the constant function $f:U\rightarrow \mathbb R$ and ...

2

One combinatorial way to do this is by using the Euler characteristic $$\chi(S^2) = V - E + F$$ as you suggested, where $V$ is the number of vertices, $E$ is the number of edges, and $F$ is the number faces in your simplicial complex. Since $\chi$ is a topological invariant, this number is the same for each triangulation, and also for each CW-complex. For a ...

2

Take any embedded annulus $A : S^1 \times [0,1] \to M_K$ such that $A(S^1 \times 0)=m$. Since $A$ is embedded, $m'=A(S^1 \times 1)$ is disjoint from $m$. We may regard $A$ as a homotopy in $M_K$ from $m$ to $m'$, and crossing with $S^1$ we get a homotopy in $S^1 \times M_K$ from $F_1 = T_m = S^1 \times m$ to the disjoint torus $F_2 = T_{m'} = S^1 \times m'$. ...

1

The fact that $$\operatorname{cor}\circ\operatorname{res}:H^n(G,A)\to H^n(G,A)$$ is multiplication by $(G:H)$ shows that the $p$-primary part of $H^n(G,A)$ embeds into $H^n(G_p,A)$, where $G_p$ is any Sylow $p$-subgroup. Since $H^n(G,A)$ is the direct sum of its $p$-primary parts, this shows that $H^n(G,A)$ is a subgroup of ...

1

I was thinking about this lying in bed this morning, and I believe that $I_G$ does contain elements $\zeta$ such that for all $E$, $\zeta_E$ is fixed by $g^*$ for all $g\in W_E$. Suppose $g\in W_E$ is not trivial. Then the following diagram commutes: $$\require{AMScd}\begin{CD} E @>{c_g}>> E \\ @VVV @VVV \\ G@>{c_g}>> G \end{CD}$$ ...

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