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Let $f\colon X \to Y$ be a map of schemes. For this to be a closed immersion, the sheaf-level requirement is that $f^\sharp\colon \mathcal{O}_Y \to f_\ast\mathcal{O}_X$ needs to be a surjective map of sheaves.

In practice we often want to check surjectivity on stalks. Now there are two possible maps on stalks, and I'm really confused whether surjectivity for the two maps is equivalent, and couldn't find any definitive source anywhere.

On the one hand, there is the usual induced map on stalks $f^\sharp_q\colon \mathcal{O}_{Y,q} \to (f_\ast\mathcal{O}_X)_q$ for all $q\in Y$. There is also the map $\phi_p\colon \mathcal{O}_{Y, f(p)} \to \mathcal{O}_{X, p}$ for all $p\in X$.

Now I understand that $f^\sharp_q$ being surjective is equivalent to $f^\sharp_q$ for all $q\in Y$. But in some proofs I've seen, people seem to assume that $\phi_p$ being surjective for all $p\in X$ is also an equivalent condition.

Question. (i) Is this second equivalence true? Does $\phi_p$ surjective implies $f^\sharp_q$ surjective, or the other way around? (ii) Is $(f_\ast\mathcal{O}_X)_q = \mathcal{O}_{X, p}$ when $f(p)=q$?

Contexts & References (feel free to ignore).

There is a related question here which refers to Qing Liu's Algebraic Geometry and Arithmetic Curves, proof of Prop 2.24, which claims (ii) of my question to be true. But I can't see why it holds, when $q\in f(X)$: I don't think the argument for this case given in that question is correct.

zh'nil
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    I think Keenan Kidwell's answer here answers your question - what do you think? – KReiser Jan 23 '24 at 00:07
  • "There is also the map $\phi_p\colon \mathcal{O}{Y, f(p)} \to \mathcal{O}{X, p}$ for all $p\in X$." - what does this sentence mean? There is a canonical map $f^{-1}(\mathcal{O}Y) \rightarrow \mathcal{O}_X$ of sheaves on $X$ and you may take the stalk at $p$ to get a map $f^{-1}(\mathcal{O}_Y)_p \rightarrow \mathcal{O}{X,p}$ – hm2020 Jan 23 '24 at 09:33
  • @KReiser Ah, thank you! I somehow hadn't come across that page. It does answer my question! - I've overlooked the topological condition that it's a homeomorphism, which will imply that (ii) is true. I will try to write this up as an answer later myself. – zh'nil Jan 23 '24 at 14:18
  • @hm2020 My map $\phi_p$ is as in when we require it to be a local homomorphism of local rings in defining a morphism of schemes (Hartshorne p.73). I think you're right that there is the map $f^{-1}\mathcal{O}Y \to \mathcal{O}_X$ by adjointness. I haven't thought about it, but it's not immediately clear to me that $(f^{-1}\mathcal{O}_Y)_p = \mathcal{O}{Y, f(p)}$, which is somewhat similar to what I'm asking in (ii). – zh'nil Jan 23 '24 at 14:24
  • I believe there is an isomorphism $f^{-1}(\mathcal{O}Y)_p \cong \mathcal{O}{Y, f(p)}$ of local rings, and composing with this morphism gives a morphism $\mathcal{O}{Y, f(p)} \rightarrow \mathcal{O}{X,p}$ as in your post. – hm2020 Jan 23 '24 at 14:29

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