Is intersection homology the usual homology of something else? Let $X$ be the sort of topological space for which it makes sense to talk about the intersection homology.  Fix a perversity $p$, or just take $p= 1/2$ if you like. 


Is there some naturally defined $X'$ such that ${}^p IH_* (X) = H_* (X')$ ?


 A: Such a space $X'$ cannot possibly depend functorially on $X$: suppose that there is a functor $F$ from the category of topological spaces (that have reasonable stratifications) and continuous maps to itself such that IH(X)=H(F(X)).  Then IH would be functorial, but considering the example 
$$(S^1 \hookrightarrow X \rightarrow S^1)=id_{S^1}$$ where $X$ is the pinched torus and $S^1$ is the non-collapsed circle (see pages 55 and 56 of Kirwan-Woolf "Introduction to Intersection Homology") would imply that the intersection homology of $S^1$ (=ordinary homology of $S^1$) imbeds in the intersection homology of $X$.  But the first IH group of $X$ (middle perversity) is $0$.
A: The short answer is "no". But you might be interested in some work Markus Banagl is doing along these lines. He has a functor that assigns to a space X an "intersection space" $I^{\bar p}X$. Then rather than studying $I^{\bar p}H_\ast(X)$, he looks at $H_\ast(I^{\bar p}X)$. However, this is not generally the same thing as $I^{\bar p}H_\ast(X)$. 
A: I'm not sure this is the kind of answer you want, but if $X$ has a small resolution $f:X' \rightarrow X$ (so $X'$ is a manifold and the dimension of fibers is sufficiently small), then there is an induced isomorphism $IH_{\ast}(X) = IH_{\ast}(X')$, and because $X'$ is smooth, the latter group is $H_{\ast}(X')$.
More precisely, a proper (birational) map is small if the set $${x \in X |  \dim f^{-1}(x) \geq r }$$ has codimension more than $2r$; such maps induce isomorphisms on IH.
Of course, there is nothing natural about $X'$, nor do small resolutions necessarily exist (see the comment below by Mike Skirvin for an easy example).  Hopefully someone more knowledgeable about IH will have something to say.
