I am unsure whether or not the following claim is true or false and whether or not my proof works or not:
Claim: Let $V \subset \mathbb{C}P^n$ be a complex $k$-dimensional, non-singular, projective algebraic variety. Then for all $i > k$ the homology groups are $H_i(V)=0$.
Is the following proof correct?
My proof: It can be shown that $V$ is induced by a $k$-dimensional manifold $\bar{V} \subset \mathbb{C}^{n+1}$, i.e. $V = q(\bar{V})$, where $q: \mathbb{C}^{n+1} \setminus \{0\} \rightarrow \mathbb{C}P^n$ is the quotient map. Because of the properties of the quotient map $q$, we can choose $\bar{V} \subset S^n$, where $S^n \subset \mathbb{C}^{n+1}$ is the $n$-sphere. From Theorem 7.1 in John Milnor's book "Morse Theory" it follows that for all $i > k$ we have $H_i(\bar{V})=0.$
Consider the map $p: V \rightarrow \bar{V}$ that sends an equivalence class $[x] \in V$ to $p(x) := \frac{x}{|x|} \in \bar{V}$. The composition of this map with $q$ is the identity on $V$, i.e. $q \circ p = id$. Hence the induced maps in homology give $q_* \circ p_* =id$, therefore: $$q_* \circ p_* =id: H_i(V) \stackrel{p_*}{\rightarrow} \underbrace{H_i(\bar{V})}_{=0 \text{, if } i>k} \stackrel{q_*}{\rightarrow} H_i(V)$$ and it follows that $H_i(V)=0$ for all $i > k$.