Let $X=\mathbb C^2$, let $X^{[n]}$ be the Hilbert scheme of length $n$ 0-cycles in $X$, and let $X^{[n]}\_0$ be the closed subscheme formed by the 0-cycles supported at 0. As far as I know $X^{[n]}\_0$ and $X^{[n]}$ have the same homotopy type. Can anybody suggest a proof? (according to Nakajima this can be proved by adapting an argument of Slodowy (Four lectures on simple groups and singularities, Section 4.3) but I am unable to do it...)



Homotopy type of Hilbert schemes of points of $\mathbb C^2$
It is a general folklore result, that if ${\mathbb C}^\times$ acts on a smooth complex variety $X$ so that the fixed point set $X^{{\mathbb C}^\times}$ is proper and the limit 
${{\rm lim}\_{\lambda \to 0} \lambda z } $
 exists for every $z\in X$, then the downward flow $$D:=\{ z\in X | {\rm lim}\_{\lambda \to \infty} \lambda z \mbox{ exists}{\}}$$ is a retract of $X$. This can be proved by Morse theory arguments as in Kirwan's book or you can prove it by first showing that the imbedding induces 


$$H\_\*(D;{\mathbb Z})\to H\_{\*}(X;{\mathbb Z})$$ an isomorphism (by induction with respect to an appropriate ordering of the set of components of $X^{{\mathbb C}^\times}$). Similarly you can show that the fundamental groups are also isomorphic. Then the relative Hurewitz theorem will tell you that they are weakly homotopy equivalent, and as they are varieties so CW-complexes, Whitehead's theorem imply that they are homotopy equivalent. 


For the Hilbert scheme you can use the ${\mathbb C}^\times$ action induced from the dilation on ${\mathbb C}^2$ mentioned above, which will have the required property, due to the fact that the Hilbert-Chow morphism is proper. 


(EDIT: more detailed discussion of this argument can now be found in Corollary 1.3.6 of the paper <http://arxiv.org/abs/1309.4914>)


