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.



Does a homeomorphism of $S^1 \times S^2$ which is homotopy to the identity has to isotope to it?
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 $GL\_n(\mathbb Z)$. 


