Do holomorphic mappings always preserve/induce Kähler forms? I was thinking in several related questions for which I haven't been able to find a proof nor a formal statement (Basically when do holomorphic mappings induce Kähler forms and how does Kählerness behave under holomorphic mappings).
So, let's suppose $X$ and $Y$ are complex manifolds. Now, I will denote by $\pi\colon X\to Y$ and $f\colon Y\to X $ some random holomorphic mappings. My questions are

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*If $(X, \omega_X)$ is Kähler, is it true that $(Y, f^*\omega_X)$ is always Kähler? (it should be true if $f$ is an immersion, but I want to know if that holds as well for other conditions, e.g., properness, submersions, etc.)

*Does the existence of a holomorphic mapping $f\colon Y\to (X, \omega_X)$ impose some restrictions on the Kählerness of $Y$ or the fibres $Y_x:= f^{-1}(x)$? (in case the latter happen to be complex manifolds as well). For example, can we find another Kähler form on $Y$ other than $f^*\omega_X$  (whenever $f^*\omega_X$ is Kähler of course)? In general, does the existence or the properties of the mapping $f$ itself impose restrictions on the the Kählerness of $Y$ and/or the fibres $Y_x$?.

*If $\pi\colon (X, \omega_X)\to Y$ is holomorphic, when can we say that $Y$ and/or the fibres $X_y:=\pi^{-1}(y)$ are Kähler as well? (and how do we construct their Kähler forms if it can be done?) [this one is particularly relavant when $\pi$ is a quotient map or a covering map]
Edit: I found this exercise (aptly labelled hard) in some online notes a few days ago. As for the weird counterxample mentioned below (a deformation of Kähler manifolds that is not a Kähler), one can trace it back to Hironaka (An Example of a Non-Kahlerian Complex-Analytic Deformation of Kahlerian Complex Structures)  

*A slightly related question: is it true that the pullback of positive forms is always a positive form, i.e., if $\omega$ is a positive $(p, p)$-form on $X$, does it hold as well for $f^*\omega$ on $Y$? Edit: This seems to be true at least for vector spaces, as you can see in page 129 of Demailly's  Complex analytic and differential geometry. 
I will appreciate if you can give me some hints/proofs/counterexamples or point me out to some solid references with fully elaborated proofs/examples, not just some yes/no kind of answers.
Regards!
 A: $1.$ This is not true. For example, if $\dim Y > \dim X$, then $f^*\omega_X$ is degenerate (in directions in the kernel of $f_*$). Even if $\dim Y = \dim X$, this issue can still occur. For example, let $Y$ be the blowup of $X$ at a point $x$ (as Ted Shifrin alluded to in the comments). If $\dim X \geq 2$, then $(f^*\omega_X)_y$ is degenerate for all $y$ with $f(y) = x$. To see this, note that for a tangent vector $v \in T_yY$ along the exceptional divisor, we have $f_*(v) = 0$, so $(f^*\omega_X)_y(v, w) = (\omega_X)_x(f_*(v), f_*(w)) = (\omega_X)_x(0, f_*(w)) = 0$ for all $w \in T_yY$.
In general, a two-form $\alpha$ is the Kähler form of a Kähler metric if and only if it is a real positive $(1, 1)$-form, see this question. As $f$ is holomorphic, $f^*\omega_X$ is a real $(1, 1)$-form, so it is the Kähler form of a Kähler metric if and only if it is positive. In the examples above, $f^*\omega_X$ is degenerate and hence not positive. If $f : Y \to X$ is a holomorphic immersion, then $f^*\omega_X$ is positive and therefore gives rise to a Kähler metric on $Y$.
$2.$ The existence of a holomorphic map $f : Y \to X$ with $X$ Kähler has no bearing on the existence of Kähler metrics on $Y$ or the fibers of $f$. One reason for this is that we always have constant maps $X \to Y$. Even if $f : Y \to X$ is non-constant, it could be the case that neither $Y$ nor the fibers of $f$ admit Kähler metrics. For example, let $Y = Y_0\times X$ where $Y_0$ is a Hopf surface and $X$ is a compact Kähler manifold, and consider the holomorphic map $f : Y \to X$ given by projection onto the second factor. The complex manifold $Y$ does not admit a Kähler metric, and neither do the fibers of the map as they are Hopf surfaces.
$3.$ The existence of a holomorphic map $\pi : X \to Y$ with $X$ Kähler does not imply that $Y$ is Kähler. For example, take $X$ and $Y$ as in point $2$ and let $\pi(x) = (y_0, x)$ where $y_0 \in Y_0$ is some chosen point. On the other hand, if $\pi^{-1}(y)$ is a complex submanifold of $X$ which is the case if $y$ is a regular value of $\pi$ for example, then the restriction of the Kähler metric on $X$ to $\pi^{-1}(y)$ is a Kähler metric. More generally, the restriction of a Kähler metric to a complex submanifold is a Kähler metric, see Proposition $3.1.10$ of Huybrechts' Complex Geometry: An Introduction for example.
If $\pi : X \to Y$ is a finite covering map and $X$ admits a Kähler metric, then so does $Y$. This is true because there is a finite covering $\rho : X' \to X$ such that the induced finite covering $\pi' : X' \to Y$ is a normal covering. Then we can average the Kähler metric $\rho^*\omega_X$ on $X'$ over the deck transformations of $\pi'$ to obtain a Kähler form which descends to $Y$.
$4.$ This is false as the counterexamples in point $1$ demonstrate (for the case of $\omega = \omega_X$ a positive $(1, 1)$-form).
