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I guess the question can be asked for all manifolds. But I am particularly interested in $S^1 \times S^2$ right now. Concrete example preferrd.

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up vote 7 down vote accepted

For $S^1 \times S^2$ I believe the answer is yes. It's a theorem of Hatcher's that the group of homeo/diffeomorphisms of $S^1\times S^2$ has the homotopy-type of

$$O_2 \times O_3 \times \Omega SO_3$$

(it's on his webpage)

I think it's easy enough to check that the subgroup of this which is homotopic to the identity is just the path-component of the identity. The issues are things like: $O_2$ and $O_3$ have two path components. $\Omega SO_3$ has two path components, so the entire group has $8$ path components. But I believe all $8$ of these components remain separated even up to only the homotopy relation.

In general of course the answer is no. For example, in the smooth category, $\pi_0 Diff(S^n)$ is a $\mathbb Z_2$-extension of the group of homotopy $(n+1)$-spheres, provided $n \geq 5$. But there are only two homotopy-equivalences of $S^n$, given by degree.

For homeomorphisms, $\pi_0 Homeo((S^1)^n)$ is a useful case. It's an observation (again of Hatcher's) that this group is an extension of $GL_n(\mathbb Z)$ by an infinite rank $2$-torsion group, provided $n \geq 5$. Note that homeomorphisms (or homotopy-equivalences) up to homotopy, for $(S^1)^n$ is just the group $GL_n(\mathbb Z)$.

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The answer is yes. The very nice paper

Hass, Joel(1-CAD); Scott, Peter(1-MI) Homotopy and isotopy in dimension three. Comment. Math. Helv. 68 (1993), no. 3, 341–364. 57N10 (57M99)

does not prove this, because this was already known at the time, but gives extensive references.

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Thank you, Igor. This is a great place to look at. – Xiaolei Wu Aug 1 '11 at 15:10

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