Let $X$ be a complex irreducible quasi-projective variety, $f:X\longrightarrow\mathbb{P}^N$ a morphism, $H\subset\mathbb{P}^N$ a hyperplane, $Z:=f^{-1}(H)$ which is irreducible, $Y\subset X$ a irriducible closed subset. Clearly we have $f(Y)\cap H=f(Y\cap Z)$, is it true that $\overline{f(Y)}\cap H=\overline{f(Y\cap Z)}$?
EDIT: Moreover $Y\cap Z$ is irreducible (hence not empty).
Thanks.
Actually, $\overline{f(Y)}\cap H$ and $\overline{f(Y\cap Z)}$ don't even have to be of the same dimension:
Let $X=Y=\mathbb A^2$ with coordinates $x,y$ and $f:X\to \mathbb P^2$ the morphism $(x,y)\mapsto [x:xy:1]$. Further let $x_0,x_1,x_2$ denote the homogenous coordinates on $\mathbb P^2$ and let $H=Z(x_0)$. Then $Z=f^{-1}H=Z(x)\subset Y=\mathbb A^2$. Now, as you observe, $f(Y\cap Z)=f(Y)\cap H=\{[0:0:1]\}$, a single (closed!) point and hence the same holds for its closure: $\overline{f(Y\cap Z)}=\{[0:0:1]\}$. At the same time, $f$ is clearly dominant, i.e., $f(Y)$ is dense in $\mathbb P^2$ and hence $\overline{f(Y)}\cap H=H$.
The problem is that quasi-projective varieties are missing some pieces. In this example if you compactify $Y$ to start with, and resolve the indeterminacies of the morphism, then the image is the entire $\mathbb P^2$ and everything is dandy. Of course, if you assumed that $Y$ was projective, then the statement would be trivially true, since in that case $f(Y)$ is closed.
To salvage the situation and get a condition for a quasi-projective variety to have this condition, you can do the following:
In addition to your setup, assume that:
$\bullet$ $f(Y)\subset \mathbb P^n$ is a quasi-projective variety. (I.e., it's open in its closure).
$\bullet$ $H\cap \overline{f(Y)}$ is irreducible (you can take this instead of assuming that $Y\cap Z$ is irreducible).
In this case it follows that $f(Y\cap Z)=f(Y)\cap H=f(Y)\cap \big(H\cap \overline{f(Y)}\big)$ is a non-empty open subset of the irreducible set $H\cap \overline{f(Y)}$ and hence it is dense in it.
The above example shows that the first condition ($f(Y)$ being quasi-projective) is necessary and Dustin's example shows that the second ($H\cap \overline{f(Y)}$ being irreducible) is. This seems to suggest that this statement is the best you can hope for. (Actually, instead of irreducibility in the second condition you could assume that $f(Y)$ intersects non-trivially all the components of $\overline{f(Y)}\cap H$, but that seems harder to check).
Even with the additional assumption in your edit, the answer is still no. Take $X=Y=\mathbb A^1$ and let $f\colon X \rightarrow \mathbb P^2$ be given by sending $x$ to $(1:x:x^2)$. Let $(a:b:c)$ denote the coordinates on $\mathbb P^2$ and take $H$ to be the hyperplane defined by $b=0$. Then $Z = f^{-1}(H)$ is the single point $x=0$ and $f(Z)$ is the single closed point $(1:0:0)$. On the other hand, $\overline {f(Y)}$ is the variety defined by $b^2=ac$, so $H \cap \overline{f(Y)}$ consists of two points $(1:0:0)$ and $(0:0:1)$.
Let $X=Y=\mathbb A^{1}$, and consider the natural inclusion $\mathbb A^1 \to \mathbb P^1$. Let $H$ be the missing point, then $Y \cap Z$ is trivial but $\overline{f(Y)}=\mathbb P^1$.
If you want to demand $Z$ nonempty, choose 3 lines, $A$, $B$, and $C$ intersecting at a point in $\mathbb P^2$. Let $X=\mathbb P^2 - A$, let $H=B$, and let $Y = X \cap C$. $f$ is just the inclusion. Then $Y \cap Z$ is again zero, but $\overline{f(Y)}=C$, which intersects $H$ at one point.
