Étale cohomology of morphism whose fibers are vector spaces Let $X\rightarrow Y$ be a morphism (may not be smooth) of varieties such that the fibres are vector spaces. Are the $l$-adic cohomologies of $X$ and $Y$ equal?
If not, under what condition (other than smoothness of the map) is sufficient to ensure that the cohomologies are equal?
What did I mean to ask?: Let $f:X\rightarrow Y$ be a continuous map of manifolds such that the fibres are vector spaces and there is a "zero section" $s:Y\rightarrow X$. Then we can actually write a homotopy equivalence between $X$ and $Y$. Therefore it follows that the cohomologies of $X$ and $Y$ are same.
I was not sure how to pose this question in the $l$-adic cohomology setup. If you assume the map is smooth then etale locally it is actually trivial, which is an easy case. Therefore i asked is there a slightly weaker condition which ensure that the cohomologies are same?
 A: Jason Starr has already given a very nice answer, and this should just be seen as a small addition to what's already been said. I'll freely use the vocabulary from Jason's answer. 
Proposition. Let $f:X \to Y$ be a smooth morphism of finite-dimensional Noetherian schemes. Suppose that for every geometric point $\overline{y} \to Y$, the fiber $f_\overline{y}$ is an etale cohomological equivalence. Then $f$ is a universal etale cohomological equivalence.
Proof. Fix $n$ invertible on $Y$. Given any $\mathcal{E} \in D^b_c(Y)=D^b_c(Y,\mathbf{Z}/n\mathbf{Z})$, write $K(\mathcal{E}) = K_Y(\mathcal{E}) = Cone(\mathcal{E} \to Rf_{\ast} f^\ast \mathcal{E})$. It's easy to check that $K(-)$ is a triangulated functor from $D^b_c(Y)$ to itself. If $g: Y' \to Y$ is any morphism, we define $K_{Y'}(-)$ analogously; note that there is a canonical base change map $g^\ast K_Y(\mathcal{E}) \to K_{Y'}(g^{\ast} \mathcal{E})$.
We now argue by induction on $\mathrm{dim}Y$ (when $\mathrm{dim}Y=0$, the result is easy).  For any fixed $\mathcal{E}$, Deligne's generic base change theorem guarantees the existence of a dense open $U \subset Y$ such that the aforementioned base change map is an isomorphism for any map $Y' \to Y$ which factors over the inclusion $U \subset Y$. Applying this for $Y'$ running over all geometric points of $U$, and using our hypothesis on the geometric fibers of $f$, we deduce that $K(\mathcal{E})|U=0$. Write $j:U \to Y$ for the evident open immersion, and let $i:Z \to Y$ be the closed complement. Looking at the usual distinguished triangle $i_* Ri^! K(\mathcal{E}) \to K(\mathcal{E}) \to Rj_* j^*K(\mathcal{E})=0$, we deduce that $K(\mathcal{E}) \cong i_* Ri^!K(\mathcal{E})$. Moreover, $i_* Ri^!K(\mathcal{E}) \cong i_* K_Z (Ri^! \mathcal{E})$, by a combination of smooth and proper base change. By hard results of Deligne, Grothendieck, and Gabber, $Ri^! \mathcal{E}$ is still bounded and constructible, so (finally) we can use induction on the dimension of the target scheme to conclude that $K_Z(Ri^! \mathcal{E})=0$, as desired.
