Let $V^k\subset \mathbb C^n$ be a sub variety, such that all its irreducible components have dimension $\ge k$. Is it true that $\mathbb C^n\setminus V^k$ has homotopy type of a CW complex of dimension $\le 2n-k-1$? 


*Comments.* 1)This is true for $k=n-1$, since in this case $\mathbb C^n\setminus V^{n-1}$ is affine. Case $k=0$ is trivial. 
2) This question would help to answer:
[An analogue of Lefschetz hyperplane theorem for complements to subvarieties in $\mathbb C^n$ ?](https://mathoverflow.net/questions/64546/an-analogue-of-lefschetz-hyperplane-theorem-for-complements-to-subvarieties-in)



Homotopy type of the complement to a subvariety of $\mathbb C^n$
Take $n=4$ and let $V = \{ z\_1=z\_2=0 \} \cup \{ z\_3=z\_4=0 \}$. I claim that $\mathbb{C}^4 \setminus V$ is homotopic to $S^3 \times S^3$, which has nontrivial homology in degree $6$, contrary to your supposed bound, which is in degree $5$.


Note that $\mathbb{C}^4 \setminus V = \left( \mathbb{C}^2 \setminus \{ (0,0) \} \right)^2$. Taking the quotient by $\mathbb{R}\_{+}$, we see that $\mathbb{C}^2 \setminus \{ (0,0) \}$ is 
homotopic to $S^3$, so $\mathbb{C}^4 \setminus V$ is homotopic to $S^3 \times S^3$. 




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I can prove the required cohomology vanishing if you require that $V$ be Cohen-Macaulay.


Write $U$ for $\mathbb{C}^4 \setminus V$. We have the Hodge-de Rham spectral sequence: $H^q(U, \Omega^p) \implies H^{p+q}(U, \mathbb{C})$. Singe $U$ is an open subset of $\mathbb{C}^n$, we have $\Omega^p \cong \mathcal{O}^{\oplus \binom{n}{p}}$ so $H^q(U, \Omega^p) \cong H^q(U, \mathcal{O})^{\bigoplus \binom{n}{p}}$.


We can identify $H^q(U, \mathcal{O})$ with a local cohomology module of $V$, which the Cohen-Macaulay condition should force to be $0$ for $q > n-k-1$. So $H^q(\Omega^p)$'s will be zero for $q>n-k-1$. Then the spectral sequence immediately forces cohomology to vanish for $p+q > n+(n-k-1)$, as you desired.


I have no idea of how to get a statement in homotopy out of the Cohen-Macaulay condition. 


