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22

As usual, there's no loss of generality in assuming that $f$ is the inclusion of a subspace $X\subset Y$, replacing $Y$ with the homotopy equivalent mapping cylinder of $f$ if necessary. By your assumptions and the five lemma, $H_*(Y,X)=0$ for $*\leq n$, and the pair $(Y,X)$ is simply connected, therefore by the Hurewicz theorems $\pi_*(Y,X)=0$ for $*\leq n$....


8

(1) The main benefit of a simplicial model category structure is explicit formulas for (co)simplicial resolutions. Remark 5.2.10 in Hovey's book identifies the functor $A\mapsto \tilde{A}^m = A\otimes \Delta[m]$ as a cosimplicial resolution (over all $m$, obviously), if $A$ is cofibrant. I needed this in my thesis, specifically Corollary 6.7 here, and was ...


8

The spin cobordism groups $\Omega^{spin}_n$ have been computed for $n \leq 127$; see section 10 of Secondary Invariants for String Bordism and tmf by Bunke and Naumann. They use MAPLE together with the decomposition of the 2-completion of $MSpin$ found by Anderson, Brown, and Peterson in their paper Spin Cobordism. In particular, one sees that $\Omega^{spin}...


8

Is there a notion of cohomology ring of X with coefficients in A? Yes, and nothing new is needed. The underlying additive group of $A$ is abelian so you take cohomology with coefficients in that abelian group; then the multiplication on $A$ is a bilinear map $A \times A \to A$ which induces a map $$H^n(X, A) \times H^m(X, A) \to H^{n+m}(X, A)$$ in the ...


6

I believe the main reasons enriched model category are simpler boils down to: Tensoring and co-tensoring by $\Delta[1]$ gives very well behaved path objects and cylinder objects adjoint to each other Slightly more generally, (co)tensoring by more general simplicial set gives an explicit construction of the (co)tensorization by spaces, and this also ...


3

Proposition 3.3 in De Turck and Gluck's "Linking Integrals in the n-sphere" states: Let $K^k$ and $L^\ell$ be disjoint closed oriented smooth submanifolds of $S^n$ with $k+\ell=n-1$ and let $f:K^k\ast L^\ell \rightarrow S^n$ be [the map sending the line segment $\{(\mathbf{x}, \mathbf{y}, u)\mid0\leq u \leq 1\}$ connecting $\mathbf{x}$ and $\mathbf{y}$ in ...


3

The answer is no. Before I explain why, let me reduce the problem to a problem in algebraic topology. Consider the natural map $Pic(X\times X)\stackrel{cycle}{\to} H_2(X\times X)\to End(H_1(X))$ where the second map sends a cycle $\sigma$ do the endomorphism $y\mapsto (\pi_2)_*(\pi_1^*y \cap \sigma)$. For the divisor $D$ as in the question, the associated ...


3

I suggest you look at Finkelsteins’s PhD thesis! L. D. Finkelstein, On the stable Homotopy of infinite loop spaces. PhD thesis. Northwestern University. 1977 The objective of this thesis is to study $QA$ when $A$ is an infinite loop space which in particular you may take $A=QX$. Also, some papers of Kuhn consider iterated application of $Q$, in particular ...


2

One way to think about this is as follows. One has a topological groupoid $\mathcal G$ with objects $G/H$ and morphisms $G \times G/H$. Similarly one has the groupoids $\mathcal G^{\prime}$, and $\mathcal H$ (with just one object $H/H$). Your bar constructions are just the classifying spaces of these three groupoids, and equivalences between these ...


1

What you call a 0-surgery on a Seifert manifold is a move that typically does not produce a Seifert manifold. It is a move that "kills the fiber" and it produces a graph manifold, homeomorphic to a connected sum of lens spaces. Since this may be a potential source of confusion, let me mention that a a 0-surgery on the $(p,q)$-torus knot is not a "0-surgery"...


1

$0$-surgery at a torus knot yields a Seifert fibered space, an explicit construction is given in Louise Moser: Elementary Surgery along a Torus Knot (PJM, 1971). Be aware that a 0-Surgery is an (1,0)-Surgery in the notation of that paper.


1

For $(\delta w) \cup v - v \cup (\delta w)$ to be a coboundary, it would need to also be a cocycle: so we would have to have $$ 0 = \delta((\delta w) \cup v - v \cup (\delta w)) = (\delta w) \cup (\delta v) - (\delta v) \cup (\delta w) $$ Let $X$ be the standard 6-simplex $[v_0,v_1,\dots,v_6]$. Define the following cochains on $X$: $$ \begin{align*} v(\...


1

No. Let $X_1=X_2=X_3=S^1$ be copies of the unit circle and consider $X_1\vee X_2\vee X_3$ with wedge basepoint $x$. Let $\gamma_i$ be a loop traversing $X_i$ whose homotopy class generates $\pi_1(X_i,x)$. Construct $X$ by attaching a 2-cell to $X_1\vee X_2\vee X_3$ by the attaching loop $\gamma_1\ast\gamma_2\ast \gamma_{3}^{-1}$. Consider an open cover $\...


1

It is independent of the choice of base point. Let $Map((X,x_0), (G_F,id))$ be the based mapping space (based at $x_0$). Let $Map(X, G_F)$ be the free mapping space. Then we have a split short exact sequence of topological groups: $$Map((X,x_0), (G_F,id)) \to Map(X, G_F) \stackrel{ev_{x_0}}{\to} G_F$$ (These are groups using pointwise multiplication). ...


1

This is not a complete answer, but rather a specific instance where the answer is yes. Let $[M] \in \mathfrak{N}_*$. Then $[M]$ contains a spin manifold if and only if all the Stiefel-Whitney numbers of $M$ involving $w_1$ or $w_2$ vanish. Here $\mathfrak{N}_* = \Omega^O_*$ is unoriented cobordism, and a a spin manifold is a manifold for which $w_1$ and $...


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