Killing homotopy groups by removing subsets

Let $X$ be a locally finite CW-complex and let $U$ be an open subset of $X$. Given a non-zero homotopy class $x\in\pi_i(U)$ say, is it possible to find a closed subset $Z\subset U$ whose removal from $U$ kills the class $x$ without creating any new homotopy classes: $$\pi_i(U\backslash Z) = \pi_i(U)/\langle x \rangle, \quad \pi_j(U\backslash Z) = \pi_j(U)\quad \mathrm{for} \quad j\neq i.$$ Is a question like this tackled in the literature somewhere?

Example: Let $X = S^1\times [0,1]$. For $U = S^1\times (0,1)$ and $x\in\pi_1(U)$ we can take $Z$ to be an embedded open interval from one end of the cylinder to the other: $Z = \{y\}\times (0,1)$ for any $y\in S^1$.

I think my class $x\in\pi_i(U)$ had better be a generator as I suspect killing $x^2$ by this method would also kill $x$. At the end of the day, I would like to find a collection of disjoint closed subsets such that if you remove them all the resulting space is contactible.

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It is not possible for $U=X=S^2$. –  Anton Petrunin Jul 26 '12 at 13:43
I think $S^2$ is fine: you can remove a single point to get a contractible space. –  Spiros Adams-Florou Jul 26 '12 at 13:56
Ah I see what you are getting at, I'm killing all homotopy groups by removing a point. So yes maybe it cannot be done one dimension at a time. –  Spiros Adams-Florou Jul 26 '12 at 14:03
Obviously closed subsets such that if you remove them, the resulting space is contractible are easy to find: just take the complement of an open ball in a top-dimensional cell. –  Will Sawin Jul 26 '12 at 14:06
Good point Will, thanks! –  Spiros Adams-Florou Jul 26 '12 at 14:14