Let $X$ be a Cohen-Macaulay scheme and $f:X\rightarrow Y$ a morphism. Under which "non trivial conditions" on $f$ can we conclude that $Y$ is also Cohen-Macaulay? By "trivial conditions" I mean smooth+surjective or étale surjective. Is it true under flatness?

### Direct summands

There are several direct summand conditions which imply this.

For example if $f$ is finite and $O_Y \to f_* O_X$ splits in the category of $O_Y$-modules, then $Y$ is also Cohen-Macaulay. The summand condition is much weaker than flatness for finite maps. This is easy, use the local cohomology criterion for Cohen-Macaulayness.

Another related condition is the following. Now suppose we are working over a field of characteristic zero. Suppose further that $X$ has rational singularities (which is Cohen-Macaulay + a little more), then if $O_Y \to R f_* O_X$ splits in the derived category, then $X$ Cohen-Macaulay (even more has rational singularities). See a paper of Sandor Kovacs in Duke.

If $X$ and $Y$ are schemes over a field (ie, equal characteristic) and $X$ is regular and $O_Y \to f_* O_X$ splits, then $Y$ is Cohen-Macaulay. This is open in mixed characteristic if I recall correctly (see a paper of Hochster and Roberts).

It is sufficient that $f$ be faithfully flat. This is a local question (I'm assuming everything is Noetherian). If you have a flat map of local rings $(R,\mathfrak{m})\to (S,\mathfrak{n})$, then we have $dim~S=dim~R+dim~S/\mathfrak{m}S$ and $depth~S=depth~R+depth~S/\mathfrak{m}S$. Since depth is bounded by dimension, this implies that $S$ is Cohen-Macaulay if and only if both $R$ and $S/\mathfrak{m}S$ are Cohen-Macaulay; cf. Bruns-Herzog, 2.1.7.