Universal property of the normalization

The affine ring statement: let $A\subseteq B\subseteq \mathrm{Frac}(A)$ with $B$ integrally closed, and let $A^{\mathrm{\prime }}$ be the integral closure of $A$ in $\mathrm{Frac}(A)$. Every $a^{\mathrm{\prime }}\in A^{\mathrm{\prime }}$ is integral over $A\subseteq B$, hence integral over $B$, and lies in $\mathrm{Frac}(A)=\mathrm{Frac}(B)$; since $B$ is integrally closed, $a^{\mathrm{\prime }}\in B$. Thus $A^{\mathrm{\prime }}\subseteq B$, and the inclusion $A^{\mathrm{\prime }}\hookrightarrow B$ is a $k$-algebra map extending $A\hookrightarrow B$.

Local construction. Let $U_i$ be a chart of $X$ and put $Y_i=f^{-1}(U_i)$, an open subvariety of the irreducible normal variety $Y$, hence normal. Read the pullback of functions in the common function fields: $f$ birational gives $k(Y)=k(X)=k(U_i)$, and $f_i=f{\mathrm{\mid }}_{Y_i}$ is dominant. For $b\in B_i\subseteq k(X)=k(Y)$ the element $b$ is integral over $A_i$; its pullback lies in $k(Y)$, and at every point $y\in Y_i$ the pullback of $A_i$ is contained in the local ring ${\mathcal{O}}_{Y,y}$ (pullback of functions regular at $f(y)$ is regular at $y$). The local ring ${\mathcal{O}}_{Y,y}$ is an integrally closed domain [F3], so $b\in {\mathcal{O}}_{Y,y}$ for every $y\in Y_i$; hence $b$ is a regular function on $Y_i$. The resulting $k$-algebra map $B_i\to {\mathcal{O}}_Y(Y_i)$ extending $A_i\to {\mathcal{O}}_Y(Y_i)$ defines, on each affine chart of $Y_i$, a morphism by the anti-equivalence [F4]; these morphisms agree by their pullback maps and glue by [F4] to a morphism $g_i\text{\hspace{0.17em}⁣}:Y_i\to U_i^{\nu }$ with $\nu \circ g_i=f{\mathrm{\mid }}_{Y_i}$.

Gluing and uniqueness. On an overlap $Y_i\cap Y_j$, the morphisms $g_i$ and $g_j$ have the same pullback on the common function field $k(X)$, because both are determined by the identifications $k[U_i^{\nu }]=B_i\subseteq k(X)=k(Y)$; hence they agree on the dense open intersection by [F4], and they glue to a morphism $g\text{\hspace{0.17em}⁣}:Y\to X^{\nu }$ over $X$ with $\nu \circ g=f$. If $g^{\mathrm{\prime }}$ is another such morphism, then $g$ and $g^{\mathrm{\prime }}$ have the same pullback on rational functions on $X^{\nu }$ (both equal the given identification $k(X^{\nu })=k(X)\to k(Y)$), so they agree on a dense open and hence everywhere by the equality lemma [F4]; this proves uniqueness. The affine statement of step 1.1 is the chartwise content of the construction, and the morphism $X^{\nu }\to X$ is the normalization by [F1].