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Let $p:Bl_0(\mathbb{A}^2) \to \mathbb{A}^2$ be blowup of plane at the origin. Is there a geometric reason why there are no sections $s: \mathbb{A^2} \to Bl_0(\mathbb{A^2})$, that is no maps to vareties with $p \circ s= id$ on affine plane $\mathbb{A^2}$.

Where I'm trying to get to is: in this comment by Zhen Lin used this recognizing the total space of the tautological line bundle $O(-1)$ on $\mathbb{P}^1$ to be identical to the blowup of affine plane at the origin.

user267839
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Here's one simple geometric reason: Consider any two distinct lines $L_1, L_2$ through the origin in $\newcommand{\AA}{\mathbb{A}}\AA^2$. The proper transforms of these two lines under the blowup at the origin are two non-intersecting lines in $\newcommand{\Bl}{\operatorname{Bl}}\Bl_0(\AA^2)$. If there were a section $s\colon \AA^2 \to \Bl_0(\AA^2)$ of the blowup map, then it would have to map each of these lines to its proper transform, because the image of a connected space under a continuous map is connected, and regular maps are continuous. But this is impossible, because this forces $s(0, 0)$ to have two different values.

Daniel Hast
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    a beautiful argument. Do you maybe know how far the original claim can be genalized? Is it in generally true that if $X$ is a variety and $Z \subset X$ a subvariety and we blow up $p:Bl_Z(X) \to X$ in $Z$ there is no section $s: X \to Bl_Z(X)$? If $X$ smooth we can work locally and use similar argument as for affine plane to show that such section not exist. But what about weird cases where eg $X$ has singular locus $Z$ and we blow it up or if $X$ don't have lines thrrough $z \in X$ and we blow up this point. Can blow ups of such pathological constructions have sections? – user267839 Nov 12 '20 at 21:50
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    Yeah, same argument should work whenever $X$ is smooth. For more general schemes $X$, it's not obvious off the top of my head why there couldn't be a section of the blowup map, though I strongly suspect the statement is true in a lot more generality than just smooth varieties. – Daniel Hast Nov 12 '20 at 23:01
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    @IsaktheXI I thought about it a bit more and was able to prove the statement for arbitrary blowups of schemes. I've written up the proof here: https://math.stackexchange.com/questions/3906612/section-of-blowup-map-of-schemes – Daniel Hast Nov 14 '20 at 00:45
  • I thank you for your effort! – user267839 Nov 16 '20 at 02:09