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If two distinct topological spaces X and Y have the same homology groups, then their symmetric products also have the same homology? Do you know if is it true? Do you Know any references for?

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3 Answers 3

As written by Dan Petersen this was proved by A. Dold. In fact the computation of the homology of the symmetric products of a topological space turns out to be difficult over the integers and over a finite field. Such computations were done by J. Milgram in for $SP^n$ in the case $n=1,2,\infty$: "The homology of symmetric products." Trans. Amer. Math. Soc. 138 1969 251–265.

J. Milgram's computations rely on results of N. Steenrod wich can be found in "Séminaire H. Cartan 1954/1955" based on the theory of constructions. Then J. Milgram uses the fact that when $X$ is a connected CW-complex of finite type $H(X)\cong H(Y)$ where $Y$ is a wedge product of Moore spaces. This result is the key point that A. Dold uses in the paper cited above and you will notice that this result is false when we consider the cohomology algebra.

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Yes, this is true. See Albrecht Dold, "Homology of symmetric products and other functors of complexes", Ann. of Math. (1958).

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A stronger statement is true. If $X$ and $Y$ are spaces such that $H_*(X) \approx H_*(Y)$ (no need to have a map that induces this isomorphism), then $SPX$ and $SPY$ are weakly homotopy equivalent. In particular if $X$ and $Y$ are CW complexes, then $SPX$ and $SPY$ are homotopy equivalent. This can be found in Hatcher: Algebraic topology, (see Hatcher's website).

More precisely:

  1. By the Dold-Thom theorem $\pi_n(SPX) \approx H_n(X).$ (Thm 4.83 in Hathcher's book.)
  2. On the other hand $SPX$ is a product of Eilenberg - Maclane spaces. (corollary 4.84 there.)

Putting these two statements together you get what I wrote.

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