We know every differential manifold can be triangulable. Let $M$ be a compact complex manifold of dimension $m$ and V be an analytic subset of dimension $s$ of $M.$ If $V$ has no singularity then $V$ is a compact complex submanifold of $M.$ Hence, V can be considered as an element of $H\_{2s}(M,\mathbb{C})$ (singular homology of M) for $V$ can be triangulable and compact. Now, consider the general case when V has singularity, as far as I know in general V is not triangulable. 


Besides, it is well-know that the analytic set $V$ has the Poincare duality $\omega$ in $H\_{DR}^{2m-2s}(M)$ (De rham cohomology of $M$), and again $\omega$ has the Poincare duality $\sigma \in H\_{2s}(M,\mathbb{C}).$ That means there exists $2s^{th}$ singular homology chain $\sigma$ such that for all 2s-form $\eta$ one has
 $$\int\_V \eta =\int\_{\sigma} \eta.$$  

Question: What is the geometric relation between $V$ and $\sigma$? On the other hand, 


**How can $V$ be considered geometrically as an element of $H\_{2s}(M,\mathbb{C})?$** 



An analytic subset as a singular homology class of a compact manifold
OK, I see. You're worried about the case when $V$ has singularities. Here are a number of things that you can do:


1. $V$ is still triangulable. This goes back Lojasiewicz, I think. So you can still represent
the fundamental class by a simplical chain as before.
2. Use currents to represent the class (cf. Griffiths- Harris pp 366-400)
3. Look up Atiyah-Hirzebruch, "Analytic cycles on complex manifolds", Topology 1, 1961
4. (My personal favourite, although some consider this overkill.) Choose a resolution of singularities $\pi\_:\tilde V\to V$, and push the fundamental class $[\tilde V]$ to $M$.


