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I am not an expert on this topic. I am trying to learn about Moore spaces of type $(G,n)$. where $G$ is abelian and $n\geq 2$.

Let $M$ be a simply connected non-compact $4$-manifold with $H_2(M;\mathbb{Z})=\mathbb{Z}^{\infty}$, i.e., the free abelian group with infinite countable rank, and $H_i(M;\mathbb{Z})=0$ for $i\neq0,2$. I have the following questions:

  1. Can I say that $M$ is a Moore space of type $(\mathbb{Z}^{\infty},2)$?

(From the paper by Milnor, "On spaces having the homotopy type of a CW-Complex", I think the answer to the above is true. However, I just want some confirmation.)

  1. Does the Moore space of the above type homotopic to the infinite wedge of $2$-spheres?

I do apologize if the questions are very silly.

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    $\begingroup$ As Cantor has recently observed, there are several infinite cardinals. You mean free abelian group of infinite countable rank? $\endgroup$
    – YCor
    Commented Oct 30, 2022 at 18:02
  • $\begingroup$ @YCor.. Yes, you are right. I have edited the question. $\endgroup$
    – gola vat
    Commented Oct 30, 2022 at 18:33
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    $\begingroup$ (1) Yes, this satisfies the definition of Moore space. (2) Yes: this has the homotopy type of a cell complex, and because the space is simply connected it is homotopy equivalent to a complex whose cell structure has only those cells demanded by its homology groups. Here, this means $M$ is homotopy equivalent to a 2-complex with one 0-cell and no 1-cells, hence an infinite wedge of 2-spheres. See Proposition 4C.1 in Hatcher's textbook on algebraic topology. $\endgroup$
    – mme
    Commented Oct 30, 2022 at 19:20

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