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Given an abelian group $G$ and an integer $n \ge 1$ we can construct a $CW$ complex such that $H_n(X) \cong G$ and $\tilde{H}_i(X)=0$ for all $i \neq n$. We call this $CW$ complex a Moore space and denote it $M(G,n)$.

Hatcher writes that an easy special case is when $G = \mathbb{Z}_m$ and we can take $X$ to be $S^n$ with a cell $e^{n+1}$ attached by a map $S^n \to S^n$ of degree $m$.

I am having a hard time working out this example for myself. In particular, how do I use this attaching map of degree $m$ to compute the homology group?

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    $\begingroup$ Use cellular homology. $\endgroup$ – Balarka Sen Dec 6 '15 at 22:27
  • $\begingroup$ Are you familiar with cellular homology, where the chain groups are free on the cells and the boundary maps are multiplication by the degree? $\endgroup$ – Eric Auld Dec 6 '15 at 22:31
  • $\begingroup$ I am somewhat familiar with cellular homology. In this example, would I take $H_{n+1}(X^{n+1}, X_n)$ to be free with basis the single copy of the $n+1$ cell we attached and the map $d_{n+1}: H_{n+1}(X^{n+1}, X_n) \to H_n(X_n, X_{n-1})$ to be the attaching map of degree $5$? $\endgroup$ – Yuugi Dec 6 '15 at 22:36
  • $\begingroup$ Related (it's not really a duplicate IMO, that other one asks for a computation with a specific CW model of $S^n$ with two cells in each dimension). $\endgroup$ – Najib Idrissi Dec 7 '15 at 17:00
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$X$ be the space obtained from attaching $D^{n+1}$ to $S^n$ by degree $m$ map. The cellular chain complex of the space is the following

$$0 \to \Bbb Z \stackrel{\times m}{\to} \Bbb Z \to 0 \to \cdots \to 0 \to \Bbb Z \to 0$$

since we have cells only at dimension $0, n$ and $n+1$, with a single cell in each of these dimensions. The chain complex has nontrivial homology only at the $n$-th level, where $H_{n}(X) \cong \Bbb Z/m\Bbb Z$. Thus, the space is an $M(\Bbb Z/m, n)$-space.


If you don't know cellular homology, use the long exact sequence for $(X, S^n)$ :$$0 \cong H_{n+1}(S^n) \to H_{n+1}(X) \to H_{n+1}(X, S^n) \stackrel{\partial}{\to} H_n(S^n) \to H_n(X) \to H_n(X, S^n) \cong 0$$

Note $H_{n+1}(X, S^n) \cong H_{n+1}(X/S^n) \cong H_{n+1}(S^{n+1}) \cong \Bbb Z$ (as $(X, S^n)$ is a CW-pair) and $H_n(S^n) \cong \Bbb Z$.

Recall that $\partial : H_{n+1}(X, A) \to H_n(A)$ sends homology class of $(n+1)$-cycles $\xi$ in $X$ relative to $A$ to homology class of the $n$-cycle $\partial \xi$ in $A$. [Hatcher, pg. $117$] Bearing that interpretation in mind, note that the generator of $H_{n+1}(X, S^n)$ corresponding to $1 \in \Bbb Z$ is represented in simplicial homology as the homology class of the relative cycle $\zeta$ in $Z^{n+1}(X, S^n)$ corresponding to a triangulation of the $(n+1)$-disk in $X$ by $(n+1)$-simplices with the sum of the faces being a cycle in $Z^n(S^n)$. This cycle in $Z^n(S^n)$ is precisely $\partial \zeta$, which is in turn represented by the degree $m$ map $S^n \to S^n$ in singular homology, as $X = D^{n+1} \cup_\varphi S^n$ where $\varphi$ is a degree $m$ map.

Thus, we conclude that $\partial$ sends $1$ to $m$. As $\partial$ is injective, $H_{n+1}(X) \cong \text{ker} \partial \cong 0$. The above sequence then boils down to the short exact sequence

$$0 \to \Bbb Z \stackrel{\times m}{\to} \Bbb Z \to H_n(X) \to 0$$

Hence, $H_n(X) \cong \Bbb Z/m\Bbb Z$. The same argument can be used to prove that $H_i(X) \cong 0$ for other $i$'s. Thus, we conclue $X$ is a $M(\Bbb Z/m\Bbb Z, n)$-space.


Someone suggested that I should provide a rigorous algebraic argument for the claim that $\partial$ is the multiplication by $m$ morphism. So here goes:

$\psi : D^{n+1} \hookrightarrow D^{n+1} \coprod S^n \to X$ be the map obtained from composing the inclusion into the disjoint union with the quotient map. $\psi|_{\partial D^{n+1}} : S^n \to S^n$ is clearly a degree $m$ map. We thus have a map of pairs $\psi : (D^{n+1}, S^n) \to (X, S^n)$. Use long exact sequence of both pairs coupled with naturality to obtain the commutative diagram

$$\require{AMScd} \begin{CD} H_{n+1}(D^{n+1}, S^n) @>>> H_n(S^n)\\ @VVV @VVV \\ H_{n+1}(X, S^n) @>>> H_n(S^n) \end{CD}$$

Note that the map $H_{n+1}(\psi) : H_{n+1}(D^{n+1}, S^n) \to H_{n+1}(X, S^n)$ can be canonically identified with the map $H_{n+1}(\bar{\psi}) : H_{n+1}(S^{n+1}) \to H_{n+1}(S^{n+1})$ where $\bar{\psi} : S^{n+1} \to S^{n+1}$ is induced from $\psi$ by quotienting the subspace $S^n$ in both spaces. One can easily see by the local degree formula that $\bar{\psi}$ is a map of degree $1$, hence $H_{n+1}(\bar{\psi})$ and thus $H_{n+1}(\psi)$ is ultimately an isomorphism. The top horizontal map is an isomorphism. The right vertical map is multiplication by $m$ because $\psi|_{\partial D^{n+1}}$ is of degree $m$. By commutativity of the diagram, the bottom horizontal map - the snake map $\partial$ - has to be multiplication by $m$ too. $\blacksquare$

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  • $\begingroup$ Hello Balarka, I've been reading through your proof, there is a sentence confusing me which is " One can easily see by the local degree formula that $\bar{\psi}$ is a map of degree 1", would you mind explaining more about that? $\endgroup$ – kousaka May 15 '16 at 20:39
  • $\begingroup$ @kousaka Local degree formula says oriented cardinality of preimage of a generic point (if it exists!) by a map between topological spaces is precisely it's degree ("sum of local degrees is global degree"). So take a generic point in $X/S^n$. That must be away from the $S^n$ you quotient. Clearly that's a "good point" because preimage of a ball around that point by $\bar{\psi}$ is also a ball inside $D^{n+1}/S^n$. And cardinality of preimage is exactly $+1$. So degree is $+1$. $\endgroup$ – Balarka Sen May 16 '16 at 3:23
  • $\begingroup$ Thanks a lot for your reply, that helps! $\endgroup$ – kousaka May 16 '16 at 12:15
  • $\begingroup$ For people who haven't seen local degree formula (like me), you can find it here: math.wisc.edu/~maxim/752notes.pdf $\endgroup$ – kousaka May 16 '16 at 12:17
  • $\begingroup$ What is $n$-skeleton of $X$? $\endgroup$ – Sigur Oct 24 '17 at 14:48

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