TITLE: Pushing-forward morphisms
QUESTION [7 upvotes]: Let $f\colon X\to Y$ be a surjective morphism of algebraic varieties, defined over an algebraically closed field.
We take a morphism $\psi\colon X\to Z$, where $Z$ in another algebraic variety, that satisfies the following condition: for all $x,x'\in X$, $f(x)=f(x')\Rightarrow \psi(x)=\psi(x')$.
This yields the existence of a unique map $\psi'\colon Y\to Z$ such that $\psi=\psi'\circ f$. What are conditions on $f$ we can put on $f$ to be sure that $\psi'$ is a morphism of algebraic varieties?
Of course, if $f$ is an isomorphism, the result is true. But is for example enough to ask that $f$ is non-ramified? Or another local condition?
I would be happy by only considering the case where $f$ is quasi-finite, even if the general case seems also interesting.

REPLY [5 votes]: If $f$ is finite surjective and unramified, I think that $\psi'$ is always a morphism. In fact we can prove that $f$ is an effective epimorphism, which means that letting $p_1,p_2:X\times_Y X\to X$ be the projections, any morphism $\psi:X\to Z$ such that $\psi p_1=\psi p_2$ factors uniquely through a morphism $\psi':Y\to Z$. Let us prove that. This is local on $Y$ so we may assume that $Y=Spec(A)$ is affine. Since $f$ is finite, then $X=Spec(B)$ with $B$ finite over $A$. Then being an effective epimorphism is equivalent to exactness of the sequence $A\to B\rightrightarrows B\otimes_A B$, which means that $A\to B$ is injective (this is ok) and that any element $b\in B$ with $b\otimes 1=1\otimes b$ in $B\otimes_A B$ comes from $A$. Now proving exactness of the sequence is étale local on $A$. Due to the étale-local structure of unramified morphisms, we can then assume that $X$ is a disjoint sum of a finite number of closed subvarities of $Y$ covering $Y$. For simplicity let us do the case of two subvarieties. Thus we have two ideals $I,J$ in $A$ such that $B=A/I\times A/J$ and $I\cap J=0$. Let $b=(a_1,a_2)\in B=A/I\times A/J$ have equal images in $B\otimes_A B$. We have $B\otimes_A B=A/I\times A/(I+J)\times A/(I+J)\times A/J$. The condition of equality means that $a_1=a_2$ mod $I+J$, hence $a_1+i=a_2+j$ for some $i\in I$, $j\in J$ and hence $b$ comes from $A$. Thus $f$ is an effective epi.
Finally let us see why this implies what you want. Since $X\to Y$ is unramified, the fibres of the maps $X\times_Y X\rightrightarrows X$ are finite sets of reduced points and checking that $\psi p_1=\psi p_2$ is just set-theoretic i.e. it just means that $\psi$ is constant on the fibres of $f$, which is your assumption.
EDIT (Dec. 31, 2015): as so often, what is left to the reader to check turns out to be 
false. Presently, the computation above works for two subvarieties but 
fails in general for more than two. The point is that if $Y$ is a union 
of $\geq 3$ subvarieties (or better closed subschemes) $X_i$ and if $X$ 
is the disjoint union of them, then the 'glueing' condition on 
$X\times_Y X$ for a map $\psi:X\to Z$ allows only to descend $\psi$ to a map defined on the scheme $Y'$ which is obtained by topological 
glueing of the $X_i$. However $Y'\ne Y$ in general. For example let $Y$ 
be the union of three lines meeting in the origin in affine plane (e.g. 
$x^3-y^3=0$ if base field has char. $\ne 3$). Let $X$ be the 
normalization. Then the glueing condition allows to descend maps 
$\psi:X\to Z$ to $Y'=$ the union of coord. axes in affine 3-space. 
However in order to descend to $Y$ one needs an extra condition which should account for the fact that any 3 nonzero tangent vectors along the 3 
components of $Y$ lie in a common plane. This condition has the 
form of a certain linear relation between the derivatives of $\psi$ at 
the 3 origins of the lines which are the components of $X$. Concretely,
the function which is equal to the coordinate on one of the three lines
of $X$ and $0$ on the other two does not descend to a function on $Y$.
Still, something from the above can be saved to produce examples of
finite unramified effective epimorphisms :

finite unramified morphisms of fibre-degree at most 2;
normalizations of ordinary singularities. Here I call ordinary
singularity a singularity that local-analytically looks like the union
of spaces $Vect(e_i,i\in I_k)$ in affine $n$-space, for some partition
$\{1\dots n\}=I_1\cup\dots\cup I_d$.

In fact I must say that I am not well aware if there is a well-established
terminology for ordinary singularities. My understanding of this notion
is that such a singularity can be reconstructed from the normalization by
topological glueing, i.e. the normalization should be finite unramified
and an effective epimorphism. But it is likely that experts in singularities will have a more relevant viewpoint.