Pushforward of semi-small maps Let $f : X \rightarrow Y$ be a semi-small map of complex projective varieties with $\dim X = n$. Apparently if we assume that $X$ is smooth, then $f_*\underline{\mathbb Q}[n]$ is perverse. (It would be much appreciated if you can provide a reference for this.) Is it still true if we only assume that $X$ is locally complete intersection? I know that $\underline{\mathbb Q}[n]$ is perverse, so I think that there is a chance.
 A: Nice to see you Sho. My expectation would be that the statement is false, but I don't have a counterexample.
As you say it is true if $X$ is smooth. More generally if $f:X \to Y$ is a semismall map, and $\mathrm{IC}_X = \mathbf Q[n]$, then $Rf_\ast \mathbf Q[n]$ is perverse. Indeed we have
$$\dim \operatorname{Supp} \mathcal H^i(Rf_\ast \mathbf Q[n]) = \dim \operatorname{Supp} R^{i+n}f_\ast \mathbf Q \leq \dim \{ y \in Y : \dim f^{-1}(y)\leq \tfrac {i+n}2 \}, $$
and the condition that $f$ is semismall says precisely that the latter quantity is at most $i$. So the support conditions are always satisfied for $Rf_\ast \mathbf Q[n]$ if $f$ is semismall. If in addition $\mathrm{IC}_X = \mathbf Q[n]$ then $\mathbf Q[n]$ is Verdier self-dual, then so is $Rf_\ast \mathbf Q[n]$, so the co-support conditions are verified too.
The point is that in the argument it was really important that $\mathbf Q[n]$ was Verdier self-dual (equivalently that $\mathbf Q[n]=\mathrm{IC}_X$). It should not be enough that $\mathbf Q[n]$ is just a perverse sheaf.
A: It seems this example https://mathoverflow.net/a/72916/89514 can be adapted to our case to produce a counterexample. Suppose $Y=\mathbb{C}^2$, $X_1$ the blow up of $\mathbb{C}^2$ at the origin, $X_2$ a copy of $X_1$, and $X=X_1\cup_Z X_2$ glued along the exceptional divisor $Z$. Then $\pi:X\to Y$ is semismall, $X$ is a locally a complete intersection because it locally looks like union of two coordinate hyperplanes   in $\mathbb{C}^3$. Then $\mathbb{C}_X$ is the extension of $\mathbb{C}_{X_1}\oplus \mathbb{C}_{X_1}$ by $\mathbb{C}_{Z}[-1]$. If $R\pi_* \mathbb{C}_X$ was perverse then $R\pi_* \mathbb{C}_Z[-1]$ would only have $H^0=ker(R\pi_* \mathbb{C}_X\to R\pi_* \mathbb{C}_{X_1}\oplus R \pi_*\mathbb{C}_{X_2})$ and $H^1=coker(R\pi_* \mathbb{C}_X\to R\pi_* \mathbb{C}_{X_1}\oplus R\pi_* \mathbb{C}_{X_2})$, whatever these are in the category of perverse sheaves, but we know that $R\pi_* \mathbb{C}_Z$ is the cohomology of $\mathbb{P}^1$, so its cohomological "width" is 2.
