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For any scheme $X$, we have a natural map $\eta_X : X \to \operatorname{Spec} \Gamma(X, \mathcal{O}_X)$, which is the unit of the global section, spectrum adjunction. I have attempted to prove that this map $\eta_X$ is an isomorphism iff $X$ is affine. My proof is below. Is it correct?

Proof: Let $f:X\to \operatorname{Spec} \Gamma(X, \mathcal{O}_{X})$ be as described above. Clearly, if $X$ is not affine, then $f$ cannot be an isomorphism, so we may move on to the converse. Thus, suppose $X$ is affine. Then $X \cong \operatorname{Spec} \Gamma(X, \mathcal{O}_{X})$ along some isomorphism, so $f \cong \operatorname{Spec} \mathrm{id}_{\Gamma(X, \mathcal{O}_{X})}$ in the arrow category. Functors preserve isomorphisms, as do isomorphisms of morphisms, so $f$ is an isomorphism (since identity maps are isomorphisms).

Brandon Harad
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    It does not at all follow that $f$ is isomorphic to $\text{Spec } \text{id}_{\Gamma(X, \mathcal{O}_X)}$ in the arrow category. This would be true if $f$ were an isomorphism, so you're assuming what you want to prove. – Qiaochu Yuan Mar 28 '25 at 03:33
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    Just wondering if you've looked at this: http://math.stackexchange.com/questions/1848370/a-scheme-is-affine-iff-the-natural-map-x-to-operatornamespec-gammax-is-an?rq=1 – aeae Mar 28 '25 at 03:35
  • @aeae I have. I was wondering if there was a way to prove the result sans the category theory, and without requiring equality of $X$ to some spectrum, which the second answer there does. – Brandon Harad Mar 28 '25 at 03:51
  • @QiaochuYuan You are correct. I do not see why $f$ being an isomorphism would imply that, but I do agree that I cannot check the commutativity diagram that would yield an arrow category morphism. Is there any way to easily fix the argument? – Brandon Harad Mar 28 '25 at 03:53
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    @BrandonHarad It suffices to check the isomorphism when X is equal to some spectrum, since $\eta$ is natural and isomorphisms satisfy 2-out-of-3. – aeae Mar 28 '25 at 04:10
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    @aeae Thank you very much, I had forgotten $\eta$ was natural. – Brandon Harad Mar 28 '25 at 06:16

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Just to summarize the discussion in the comments:

Let $X$ be a scheme and $i:X\to \text{Spec}(R)$ an isomorphism of locally ringed spaces, then by naturality of $\eta$ the following diagram commutes

$$\require{AMScd} \begin{CD} X @>{\eta_X}>> \text{Spec}(\Gamma(X,\mathcal{O}_X)\\ @V{i}VV @VV{\text{Spec}(\Gamma(i))}V \\ \text{Spec}(R) @>>{\eta_{\text{Spec}(R)}}> \text{Spec}(\Gamma(\text{Spec}(R),\mathcal{O}_{\text{Spec}(R)}) \end{CD}$$

Since $i$ and $\text{Spec}(\Gamma(i))$ are both isomorphisms, $\eta_X$ is an isomorphism if and only if $\eta_{\text{Spec}(R)}$ is an isomorphism by the 2-out-of-3 property. So it suffices to check $\eta_X$ is an isomorphism when $X=\text{Spec}(R)$, in which case it follows from @Claudius 's answer https://math.stackexchange.com/a/1849439/1358377.

aeae
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