Suppose $f:X\to Y$ is a finite morphism of varieties and $\mathcal{F}$ is a CohenMacaulay sheaf on $Y$. Under what conditions on $f$ is $f^*\mathcal{F}$ CohenMacaulay?

3$\begingroup$ The only reasonable condition is if $f$ is flat. $\endgroup$ – Mohan Dec 15 '17 at 1:34
As far as being true for the most general situation, you need: $X,Y$ to be CohenMacaulay, $\dim \mathcal F_x =\dim Y_x$ ($\mathcal F_x$ is maximal CohenMacaulay) for all $x$ in the support of $\mathcal F$, and $f$ to have finite flat dimension. (more is true with a little extra technical assumption, you need the ``CohenMacaulay defect" to be constant along $f$, see the last paragraph).
Then what you need follows from the two local statements about f.g modules over a local ring $R$.
1) If M is maximal CM, and N has finite projective dimension, then $Tor_i(M,N)=0$ for all $i>0$ (M, N are Torindependent). See: http://www.math.lsa.umich.edu/~hochster/711F06/L11.20.pdf
2) If $M,N$ are Torindependent and $pd_RN<\infty$, we also have the depth formula: $depth(M\otimes N) + depth(R) = depth(M)+depth(N)$ See: https://www.math.unl.edu/~siyengar2/Papers/Abform.pdf
In particular, if N is CM then $M\otimes N$ is also CM of same depth. (here we use that $R$ is CM, so $depth(R) =depth(M)$).
If you drop any condition, it is not hard to find examples to show that the statement is no longer true. On the other hand, for special $X,Y,\mathcal F$, sometimes one can say a bit more. For example, if $R$ is a complete intersection, Torindependence forces the depth formula to hold, without knowing that $N$ has finite flat dimension.
Added in response to OP's request: By looking at the depth formula in 2), the following more technical, but general statement is true: suppose $f: R\to S$ is a finite, local map of finite flat dimension, and $dim(R)depth(R)=dim(S)depth(S)$. Then for a maximal CM module $M$ over $R$, $M\otimes_R S$ is (maximal) CM over $S$. This cover both cases when $R,S$ are CM ( both sides of the equality is $0$), or if $f$ is flat (both dim and depth are preserved).

$\begingroup$ Thanks for the detailed answer! What can we say if $X$ is not CM? For example, if $f$ is flat, it seems $f^*F$ is automatically CM, without conditions on $X$ or $Y$. Can we do better? $\endgroup$ – Lucas MasonBrown Dec 15 '17 at 13:03

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