Section of blowup map of schemes Let $X$ be a scheme. Let $\mathcal{I} \subseteq \mathcal{O}_X$ be a quasi-coherent sheaf of ideals on $X$, and let $Z \subseteq X$ be the closed subscheme corresponding to $\mathcal{I}$. Let $X' := \operatorname{Bl}_Z(X)$ be the blowup, and let $b\colon X' \to X$ be the blowup morphism. Suppose there exists a section of $b$, that is, a morphism $s\colon X \to X'$ such that $b \circ s = \operatorname{id}_X$. Is it necessarily true that $b$ is an isomorphism?
(This is a generalization of this question that was asked in the comments; I decided to make this a separate question since it goes beyond the scope of the original question. I initially thought I only had a partial answer, but in the process of writing it up, I realized it works in full generality, so I've provided an answer myself.)
 A: Yes, if the blowup morphism $b$ has a section, then $b$ is an isomorphism.
By Stacks tag 01O2, the blowup morphism $b$ is separated. By Stacks tag 01KT, the section $s$ is a closed immersion. Since $s$ is a closed immersion, the image $s(X)$ is closed. By Stacks tag 02OS, the blowup morphism $b$ restricts to an isomorphism on the open subscheme $b^{-1}(X \setminus Z)$, so $b^{-1}(X \setminus Z) \subseteq s(X)$. Since $s(X)$ is closed, we in fact have $\overline{b^{-1}(X \setminus Z)} \subseteq s(X)$.
In particular, $s(X)$ contains all irreducible components of $X'$ that have nonempty intersection with $b^{-1}(X \setminus Z)$, that is, all irreducible components of $X'$ that are not contained in $b^{-1}(Z)$. By Stacks tag 0BFM, no irreducible components of $X'$ are contained in $b^{-1}(Z)$, so $s(X) = X'$. That is, $s$ is surjective. In particular, $s$ is a homeomorphism with continuous inverse $b$.
By Stacks tag 02OS, the exceptional divisor $E = b^{-1}(Z)$ is an effective Cartier divisor on $X'$. We show that $Z = s^{-1}(E)$ is an effective Cartier divisor on $X$. This is a local question.
By Stacks tag 0804 and the definition of effective Cartier divisor, we reduce to the following local situation: we have a commutative ring $A$, an ideal $I \subseteq A$, an element $a \in I$, and a surjective map $\varphi\colon A[\frac{I}{a}] \to A$ such that the composition $A \overset{i}{\to} A[\frac{I}{a}] \overset{\varphi}{\to} A$ is the identity, where $A[\frac{I}{a}]$ denotes the affine blowup algebra and $i(x) = x/a^0$ for all $x \in A$. We want to show that, for each non-zerodivisor $f \in A[\frac{I}{a}]$, the element $\varphi(f)$ is also a non-zerodivisor.
Write $f = x/a^n$ with $x \in I^n$. Let $z \in A$ such that $\varphi(f) z = 0$. Let $z' = y/a^m \in A[\frac{I}{a}]$ such that $\varphi(z') = z$ and $y \in I^m$. Then
$$xy = \varphi(i(x) i(y)) = \varphi(a^{m+n} fz') = \varphi(a^{m+n}) \varphi(f) \varphi(z') = a^{m+n} \varphi(f) z = 0,$$
so $fz' = (xy)/a^{m+n} = 0$. Since $f$ is a non-zerodivisor, this implies $z' = 0$, so $z = 0$ and $\varphi(f)$ is also a non-zerodivisor.
Thus $Z$ is an effective Cartier divisor on $X$. By the universal property of blowing up, $b$ is an isomorphism.
