Theorem 76.36.5 (Stein factorization; general case). Let $S$ be a scheme. Let $f : X \to Y$ be a proper morphism of algebraic spaces over $S$. There exists a factorization
\[ \xymatrix{ X \ar[rr]_{f'} \ar[rd]_ f & & Y' \ar[dl]^\pi \\ & Y & } \]
with the following properties:
the morphism $f'$ is proper with connected geometric fibres,
the morphism $\pi : Y' \to Y$ is integral,
we have $f'_*\mathcal{O}_ X = \mathcal{O}_{Y'}$,
we have $Y' = \underline{\mathop{\mathrm{Spec}}}_ Y(f_*\mathcal{O}_ X)$, and
$Y'$ is the normalization of $Y$ in $X$ (Morphisms of Spaces, Definition 67.48.3).
Proof.
We may apply Lemma 76.36.1 to get the morphism $f' : X \to Y'$. Note that besides the conclusions of Lemma 76.36.1 we also have that $f'$ is separated (Morphisms of Spaces, Lemma 67.4.10) and finite type (Morphisms of Spaces, Lemma 67.23.6). Hence $f'$ is proper. At this point we have proved all of the statements except for the statement that $f'$ has connected geometric fibres.
It is clear from the discussion that we may replace $Y$ by $Y'$. Then $f : X \to Y$ is proper and $f_*\mathcal{O}_ X = \mathcal{O}_ Y$. Note that these conditions are preserved under flat base change (Morphisms of Spaces, Lemma 67.40.3 and Cohomology of Spaces, Lemma 69.11.2). Let $\overline{y}$ be a geometric point of $Y$. By Lemma 76.36.3 and the remark just made we reduce to the case where $Y$ is a scheme, $y \in Y$ is a point, $f : X \to Y$ is a proper algebraic space over $Y$ with $f_*\mathcal{O}_ X = \mathcal{O}_ Y$, and we have to show the fibre $X_ y$ is connected. Replacing $Y$ by an affine neighbourhood of $y$ we may assume that $Y = \mathop{\mathrm{Spec}}(R)$ is affine. Then $f_*\mathcal{O}_ X = \mathcal{O}_ Y$ signifies that the ring map $R \to \Gamma (X, \mathcal{O}_ X)$ is bijective.
By Limits of Spaces, Lemma 70.12.2 we can write $(X \to Y) = \mathop{\mathrm{lim}}\nolimits (X_ i \to Y_ i)$ with $X_ i \to Y_ i$ proper and of finite presentation and $Y_ i$ Noetherian. For $i$ large enough $Y_ i$ is affine (Limits of Spaces, Lemma 70.5.10). Say $Y_ i = \mathop{\mathrm{Spec}}(R_ i)$. Let $R'_ i = \Gamma (X_ i, \mathcal{O}_{X_ i})$. Observe that we have ring maps $R_ i \to R_ i' \to R$. Namely, we have the first because $X_ i$ is an algebraic space over $R_ i$ and the second because we have $X \to X_ i$ and $R = \Gamma (X, \mathcal{O}_ X)$. Note that $R = \mathop{\mathrm{colim}}\nolimits R'_ i$ by Limits of Spaces, Lemma 70.5.6. Then
\[ \xymatrix{ X \ar[d] \ar[r] & X_ i \ar[d] \\ Y \ar[r] & Y'_ i \ar[r] & Y_ i } \]
is commutative with $Y'_ i = \mathop{\mathrm{Spec}}(R'_ i)$. Let $y'_ i \in Y'_ i$ be the image of $y$. We have $X_ y = \mathop{\mathrm{lim}}\nolimits X_{i, y'_ i}$ because $X = \mathop{\mathrm{lim}}\nolimits X_ i$, $Y = \mathop{\mathrm{lim}}\nolimits Y'_ i$, and $\kappa (y) = \mathop{\mathrm{colim}}\nolimits \kappa (y'_ i)$. Now let $X_ y = U \amalg V$ with $U$ and $V$ open and closed. Then $U, V$ are the inverse images of opens $U_ i, V_ i$ in $X_{i, y'_ i}$ (Limits of Spaces, Lemma 70.5.7). By Theorem 76.36.4 the fibres of $X_ i \to Y'_ i$ are connected, hence either $U$ or $V$ is empty. This finishes the proof.
$\square$
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