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2

It sounds like the question arises from two points of confusion. 1) If you think of * as an element of $X_0$, then you need to remember that an element of $X_0$ gives rise, via the degeneracy maps, to an element in each $X_n$. (Play around with the simplicial identities to see why you get a well-defined element in $X_n$, regardless of which degeneracy maps ...


0

Could you expand on your motivation for this question? I think I have found a counterexample, essentially because if this identity was true, it would have to be an easy consequence of the simplicial identities. The functor $F$ that sends a simplicial set $K_*$ to the $n$-simplices satisfying this condition is corepresented by some simplicial set $X^n$, ...


1

It's exactly as if $x_k$ were the $k$th face of some $n$-simplex $y$ (i.e. $x_k = d_k y$), except that the full simplex $y$ may not necessarily exist. As Zhen Lin suggest in the comments, drawing a picture is quite helpful. For example when $n = 2$ and $i = 0$, this looks like this: For $n=3$ it looks like a (hollow) tetrahedron with the face opposite to ...


1

As explained in the other answer, if $(X, A)$ is a CW-pair, i.e., $X$ is a CW-complex and $A$ is a subcomplex, then one has the long exact sequence of homology groups $$\cdots \longrightarrow \tilde{H}_{k-1}(X/A) \longrightarrow \tilde{H}_k(A) \stackrel{i_*}\longrightarrow \tilde{H}_k(X) \stackrel{q_*} \longrightarrow \tilde{H}_k(X/A) \longrightarrow ...


2

What you wrote is in general completely false. Consider $A = [0,1]$ and $B = \{0,1\}$. Then $H_1(A) = 0 = H_1(B)$, but $A/B$ is a circle and $H_1(A/B) = \mathbb{Z}$. In general what is true is that if the inclusion $B \subset A$ is a cofibration (for example if $B$ is a sub-CW-complex of $A$), then the reduced homology group $\tilde{H}_k(A/B)$ is isomorphic ...


3

Here's a Mayer-Vietoris argument. Write $X$ as a union $\cup_{i=0}^\infty T_i$ where each $T_i$ is a torus with two boundary components (except $T_0$, which has only one boundary component). Let $A= \sqcup_{i=0}^\infty T_{2i}$ and $B= \sqcup_{i=0}^\infty T_{2i+1}$. Then $A \cap B$ is a countably infinite disjoint union of circles $\sqcup_{i=1}^\infty ...


4

Here is an answer without Mayer–Vietoris – maybe it's possible to use the technique you describe, but the following approach seems easier to me. Homology commutes with (some) directed colimits. You can e.g. find a precise statement in Hatcher's book Algebraic Topology, Proposition 3.33. Your space $X$ can naturally be seen as a directed colimit ...


1

An answer to the second question is indeed known, and is discussed for example in a paper by Eduardo Hoefel, Muriel Livernet, and Jim Stasheff as Definition 2.9: http://arxiv.org/abs/1312.7155. Briefly, suppose $M$ is an $A_\infty$-space and has an $A_\infty$-action on $X$. Then the coherences induce maps $$K_{n+1} \times M^n \to M$$ and $$K_{n+1} \times ...


0

Pursuant to Qiaochu's hint, presheaves aren't essential to this part of the argument, which follows from the same argument for simplicial sets. Specifically, given a simplicial set $S_\bullet$ and a set $T$, maps $\psi\colon\pi_0 S \to T$ correspond to maps $S_\bullet \to T$ in a way following from the simplicial identities, specifically that $$d_i d_j = ...


1

There is no such algorithm. By a famous theorem of Markov, there is in fact no algorithm that takes two triangulations of compact $4$-manifolds and decides whether the manifolds are homeomorphic (you can replace $4$ with any $n\geq 4$ here as well). The idea is that given a finite presentation of a group, you can computably describe a $4$-manifold with ...



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