If $f: X\to Y$ is a separable map between varieties that is a bijection on closed points, is it true that $f$ remains separable when restricted to an integral subscheme $Z\subset X$?


If we drop the condition that $f$ is a bijection, the only relevant example I know is the quasielliptic fibration $\{y^2=x^3-t\}\to \mathbb{A}^1$ that maps $(x,y,t)$ to $t$, where we are working over a field of characteristic 3. If we restrict to the locus $y=0$, then we get the Frobenius map $\mathbb{A}^1\to \mathbb{A}^1$. 



Restriction of separable map
Let $k$ be an algebraically closed field of characteristic $2$. Consider $$Y=\text{Spec}\ k[t,v,w]/\langle w^2+v^3+tv^2\rangle, \ \ X=\text{Spec}\ k[t,u].$$ Let $f$ be the morphism determined by the $k$-algebra homomorphism $$k[t,v,w]/\langle w^2 +v^3+tv^2 \rangle \to k[t,u], \ \ v\mapsto u^2+t, \ \ w\mapsto u(u^2+t).$$ This $k$-algebra homomorphism is the normalization. Thus $f$ restricts to an isomorphism on the smooth locus $D(u^2+t)\to D(v)$. 


For the integral closed subscheme that equals the singular locus, namely $\text{Zero}(v,w)$, the restriction of $f$ corresponds to the $k$-algebra homomorphism $$k[t,v,w]/\langle v,w,w^2+v^3+tv^2 \rangle \mapsto k[t,u]/\langle u^2+t\rangle,$$ that is,$$ k[t]\mapsto k[u], \ \ t\mapsto u^2.$$ This is the Frobenius morphism on the affine line over $\text{Spec}\ k$. Thus, this morphism is a bijection on closed points, but it is also purely inseparable.


