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I've been reading Hatcher's survey "A Short Exposition of the Madsen-Weiss Theorem". In it, he discusses the Barratt-Priddy-Quillen theorem, which says that the homology of the infinite symmetric group is the same as the homology of one component of $\Omega^{\infty} S^{\infty}$. My question : what space exactly is being referred to in this notation? It would appear that he intends to take the limit of $\Omega^n S^n$ as $n$ goes to infinity; however, I don't see a natural inclusion $\Omega^n S^n \subset \Omega^{n+1} S^{n+1}$, so I don't know how to make sense of this. Another possible interpretation would be to take the limit of $\Omega^n S^{\infty}$ as $n$ goes to infinity, but again I don't know a natural inclusion $\Omega^n S^{\infty} \subset \Omega^{n+1} S^{\infty}$.

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up vote 15 down vote accepted

Your first statement is correct: it is taking a colimit of $\Omega^n S^n$. The inclusion from one to the next is called the "suspension" map. The points of $\Omega^n S^n$ are basepoint-preserving functions $S^n \to S^n$, and the suspension map takes such a function $f$ and sends it to $f \wedge id: S^n \wedge S^1 \to S^n \wedge S^1$.

This is easier to describe in terms of another model. $S^n$ is homeomorphic to the space obtained by taking $[0,1]^n$ and identifying the boundary to a single point. Under this identification, you could describe a function $S^n \to S^n$ using coordinates as a function $$(x_1,\ldots,x_n) \mapsto f(x_1,\ldots,x_n),$$ and the suspension map sends this to the function $$(x_1,\ldots,x_n,x_{n+1}) \mapsto (f(x_1,\ldots,x_n), x_{n+1})$$ Of course, you have to check that this preserves the equivalence relation if $f$ does and so on.

Most models of $S^\infty$ are contractible, and hence so are $\Omega^n S^\infty$; these are definitely not what you want. The notation is not to be taken literally; at best, $S^\infty$ should denote an "infinite suspension" operator which doesn't actually produce spaces as output.

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Thanks! This is very helpful! – George Oct 27 '11 at 5:00

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