Let $X\subseteq\mathbb{C}P^n$ be a smooth subvariety. Let $C\subseteq X$ be an algebraic curve
and let $[C]\in H\_2(X;\mathbb{R})$ be the homology class it determines.
Does there exist a continuous map $u:\Sigma\longrightarrow X$ satisfying properties (1),(2) and (3) below?
(1) $\Sigma$ is a genus zero nodal curve (that is, a tree of spheres);
(2) $u$ is holomorphic on each irreducible component of $\Sigma$;
(3) $u\_\*[\Sigma]=[C]$;


Thank you.



homology class of an algebraic curve versus homology class of a genus zero holomorphic curve
A complex analytic space $X$ is called Brody hyperbolic if any holomorphic map $\mathbb C\to X$ is constant. There are many spaces that are Brody hyperbolic ([see this paper for instance](http://arxiv.org/abs/0902.3741)) and thus what you ask will not happen. (In fact, according to Kobayashi's conjecture all general hyperssurfaces of high enough degree are Brody hyperbolic). 


In fact, this is way too much firepower for that. It seems to me that essentially what you are asking is whether the cone of effective curves is generated by rational curves over $\mathbb Z$. Even if you ask this over $\mathbb Q$ it does not happen often, even when there are plenty of rational curves around. Take for instance a variety that is Brody hyperbolic (say a high genus curve) and its product with a rational curve. You will never get a rational curve "bend" towards the other direction.


One case when this actually happens is if $X$ is Fano. Then by *the* Cone Theorem the cone of effective curves is generated by rational curves. Actually, I am not sure if you can do it with a connected $\Sigma$, but you can certainly do it with the disjoint union of a few spheres. However, you have to allow rational coefficients, but you only need finitely many actual maps. It's all in the coefficients.


