For a smooth complete variety $X$ over a field $F$, we say that $X$ is incompressible if every rational map from $X$ to $X$ is dominant. If $X$ is a smooth complete variety of dimension $d$ over $F$, let’s say that $X$ satisfies property (P) if for every cycle $\alpha\in \text{CH}^{d}(X\times X)$, then $(p\_1)\_{\*}(\alpha)=1$ implies $(p\_2)\_{\*}(\alpha)\neq 0$, where $p\_i$ are the relevant projections, and $\text{CH}^0(X)$ is identified with $\mathbb{Z}$.


It is not hard to show that if $X$ has property (P), then $X$ is incompressible. Is the converse true? I would guess not because perhaps one could have some incompressible $X$ with a rational map giving rise to a cycle with pushforward along the first projection giving the cycle 1, and pushforward along the second giving the cycle 2, which we could then use to construct a cycle showing $X$ does not satisfy property (P), but I don’t know any example of an $X$ with these properties.



Cycles on an incompressible variety
The answer is no, even for $d=1$. Note that for curves over a field $F$ (edit: geometrically connected ones, so that global sections of the curve are just $X$), incompressibility is equivalent to there being no $F$-rational points. On the other hand, there exist curves with no $F$-rational points (hence incompressible) which still admit divisors defined over $F$ of degree $1$ - for instance, any curve over a finite field with no rational points will work (see [here](https://math.stackexchange.com/q/200520/127263)).


So let $X$ be an incompressible curve which admits a divisor $D$ of degree $1$, and consider the cycle $\alpha=X\times D\in\mathrm{CH}^1(X\times X)$. Its pushforward along $p\_1$ is going to be equal to $\deg(D)X$ which is $1$ under your identifications, but since all the components of $\alpha$ are "horizontal", the pushforward along the other projection will vanish.


