How to get a Stein space which has homotopy type of suspension of another Stein space Let $V^n$ be a Stein space(or Stein manifold) in $\mathbb{C}^N$. I want to construct a Stein space(or Stein manifold) $W^{n+1}$ such that $H_i(V;\mathbb{Z})=H_{i+1}(W; \mathbb{Z}).$
If we take the suspension of $V,$ is it a Stein space? Or can I get a Stein space $W^{n+1}$ which has the same homotopy as the suspension of $V$?
 A: I'm not sure what you mean by "suspension" of $V$ here. The notion of suspension I have in mind (doubling the cone of $V$ over its base) doesn't yield a manifold, and even if it did it would give an odd-dimensional manifold, so the answer to your question would be no.
About the homotopy type, the answer is yes. Since an Stein $n$-manifold has the homotopy type of an $n$-dimensional cell complex and suspension only increases dimension by 1, it's enough to know that every cell complex of dimension at most $n$ there is a (complex) $n$-dimensional Stein manifold with the same homotopy type. This was proved by Eliashberg: Stein manifolds (up to deformation) can be described by (Legendrian) handlebody diagrams, and you can get infinitely many of those thickening a cell complex to a handle decomposition. (This can be done in an essentially arbitrary way if $n>2$, and with a little bit of care when $n=2$.)
The general case is contained in Eliashberg's paper (Topological characterization of Stein manifolds of dimension > 2, Int. Math. J. 1, 1990), and is discussed in Cieliebak and Eliashberg's From Stein to Weinstein and back.
The case of $n = 2$ is mentioned in the two previous references, and is studied in depth in Gompf's Handlebody construction of Stein surfaces (Ann. Math. (2) 148, no. 2, 1998); you can also look at Ozbagci and Stipsicz's book Surgery on contact 3-manifolds and Stein surfaces.
