Let $f:X \to Y$ be a holomorphic map between two smooth complex projective manifolds. Is there a good notion of pullback of algebraic cycles by $f$ which preserves degree in the following sense: Suppose $V_1, V_2$ be two cycles in $Y$ with the same Hilbert polynomial and suppose $f^{1}(V_i)$ are algebraic varieties. Then will the pullbacks of $V_i$ by $f$ have the same dimension and degree (as algebraic cycles)? A small comparison with the divisor case is as follows: As far as I understand that pullback of line bundles preserves degree in the sense as above. By identifying Picard group with the divisor class group we have a similar statement as above.

Yes, as long as $f^* \mathcal O(1)$ is a rational mutliple of $\mathcal O(1)$. However, we will preserve not the dimension, but the codimension, as suggested by the case of divisors. We use the standard notion of pullback of cycles in algebraic geometry (e.g. Hartshorne appendix $A$). This behaves well with respect to the intersection pairing, so $f^* A \cdot f^* B = f^*( A \cdot B)$. Next recall that the degree of a cycle $V$ of dimension $d$ is just $V \cdot H^d$, where $H$ is the hyperplance class. Assume $H_Y = q f^* H_X$, and let $k$ be the relative dimension of $f$. Then the pullback of $V$ has dimension $k+d$, so $f^* V \cdot H_X^{k+d} = f^* V_1 \cdot H_X^d \cdot H_X^k = q^d f^* V_1 \cdot f^* H_Y^d \cdot H_X^k= q^d f^* (V_1 \cdot H_Y^d) \cdot H_X^k$ but $V_1 \cdot H_Y^d$ is just the class of $n$ points, where $n$ is the degree of $V$, so the degree of $f^* V$ is the degree of $V$ times $q^d f^* (pt) \cdot H_X^k$. 

