# Is geometric realization of the total singular complex of a space homotopy equivalent to the space?

Let $X$ be a topological space and let $|Sing(X)|$ be the geometric realization of the total singular complex of $X$.

Then $|Sing(X)|$ is a CW complex with one cell for each non-degenerate singular simplex. There's a natural map $f:|Sing(X)|\to X$ and there's a theorem that says that $f$ is a weak homotopy equivalence. That is, $f$ induces isomorphisms of homotopy groups.

Then it seems that Whitehead theorem applies and gives that $f$ is homotopy equivalence as long as $X$ is homotopy equivalent to a CW complex (i.e. $X$ is m-cofibrant). Is that correct?

Is there an example when $f$ is not homotopy equivalence? Any examples that come up in "real life"?

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The map from the (realization of the) singular complex of a space $X$ to $X$ is a homotopy equivalence if and only if $X$ is homotopy equivalent to a CW complex, so to get examples where the map is not a homotopy equivalence you just need spaces that are not homotopy equivalent to CW complexes. There are plenty of these that come up all the time in various places outside algebraic topology. For example, a compact space with infinitely many path-components isn't homotopy equivalent to a CW complex. For an explicit example, take the Cantor set, or more simply just a convergent sequence with its limit point. Suspending these examples gives path-connected examples, and iterated suspensions give examples with higher connectivity. More generally, any compact space having a nonfinitely generated homology group is an example. There are lots of other ways that spaces not homotopy equivalent to CW complexes arise too.

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Allen, thank you very much! I realized that my question was rather stupid. If X is not homotopy equivalent to a CW complex then X is not equivalent to |SX|! Thank you for your examples! – Zarathustra May 22 '10 at 3:05