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Suppose we're working over an algebraically closed field $k$. If $V\subseteq\mathbb{A}^n$ and $W\subseteq\mathbb{A}^m$ are affine algebraic sets, there is a well known bijective correspondence $$ \operatorname{Mor}_\mathrm{reg}(V,W)\leftrightarrow\operatorname{Hom}_{k-\mathrm{alg}}(k[W],k[V]). $$

I assume this fails in general if $V$ and $W$ are not affine algebraic sets. Consider the case where $V=\{*\}$ is a point, and $W=\mathbb{A}^2\setminus\{(0,0)\}$. So $V$ is affine, but it's also a well known fact that $W$ is not affine. My question is, why does the natural map

$$ \operatorname{Mor}_{\mathrm{reg}}(V,W)\to\operatorname{Hom}_{k-\mathrm{alg}}(k[W],k[V]) $$

fail to be surjective, so that the bijection does not hold in this case?

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    $\begingroup$ The problem is that for $V$ and $W$ not affine, $k[V]$ (many times) does not give you much information about $V$. For example, for any projective variety $V$, $k[V]=k$. $\endgroup$ – rfauffar Nov 7 '14 at 12:53
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This is similar to Hoot's answer, but since you wanted more detail, I thought I'd add it.

I tried to translate this result into classical language the best I could. Hopefully it makes some sense to you!


There is a very nice theorem, called algebraic Hartog's lemma, which says the following (in classical language)[you can find the proof in Vakil, or Qing Liu]

Theorem(Algebraic Hartog's Lemma): Let $X$ be a normal irreducible variety of dimension $n$, and let $Z\subseteq X$ be a variety of dimension less than or equal to $n-2$. Then, the restriction map $k[X]\to k[U]$ is an isomorphism, where $U=X-Z$.

The algebraic version of this theorem says that if $A$ is a normal domain, then

$$A=\bigcap_{\text{ht}(\mathfrak{p})=1}A_\mathfrak{p}$$

where the intersection takes place in $\text{Frac}(A)$. Intuitively, it says that since the zero set of a function has codimension 1, any open subset, which is the complement of a set of codimension at least $2$, can't invert anything. More symbolically, if $\displaystyle \frac{f}{g}$ is to be a function on $X-Z$, you need that the vanishing set $V(g)$ of $g$ is contained in $Z$. But, $V(g)$ has dimension $n-1$, so if $Z$ has dimension less than or equal to $n-2$, you can't have $V(g)\subseteq Z$. So, you can't invert anything.

In particular, this says that if we consider the inclusion $\mathbb{A}^2-\{(0,0)\}\to \mathbb{A}^2$, then the induced map of rings of functions $k[\mathbb{A}^2]\to k[\mathbb{A}^2-\{(0,0)\}]$ is an isomorphism of $k$-algebras. So, now, assume that the map

$$\text{Mor}(\ast,\mathbb{A}^2-\{(0,0)\})\to \text{Mor}(k[\mathbb{A}^2-\{(0,0)\}],k)$$

were a bijection. Then, consider the following commutative square

$$\begin{matrix}\text{Mor}(\ast,\mathbb{A}^2-\{(0,0)\}) & \to & \text{Mor}(k[\mathbb{A}^2-\{(0,0)\}],k)\\ \downarrow & & \uparrow\\ \text{Mor}(\ast,\mathbb{A}^2) & \to & \text{Mor}(k[\mathbb{A}^2],k)\end{matrix}$$

where the left hand vertical map comes from the inclusion $\mathbb{A}^2-\{(0,0)\}\to\mathbb{A}^2$, and the right hand side comes from the restriction map $k[\mathbb{A}^2-\{(0,0)\}]\to k[\mathbb{A}^2]$.

Now, since $\mathbb{A}^2$ is affine, the bottom horizontal arrow is a bijection. By algebraic Hartog's lemma, the right vertical map is a bijection. So, if we assume that the top horizontal arrow is a bijection, then we must conclude that the left vertical map is a bijection--but it's not!

The key to all of the above was not really just that $k[\mathbb{A}^2]\cong k[\mathbb{A}^2-\{(0,0)\}]$, but that (by algebraic Hartog's lemma) it was the restriction map that induced the isomorphism of $k$-algebras.

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  • $\begingroup$ Dear Alex, thanks for the answer! This is more clear to me. $\endgroup$ – Jacqueline Pauwels Nov 24 '14 at 3:11
  • $\begingroup$ @JacquelinePauwels No problem! Let me know if there is anything I can clarify. :) $\endgroup$ – Alex Youcis Nov 24 '14 at 3:13
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Here's one way to think about the problem, which will probably lead you to an answer even if the details aren't immediately clear. $$\begin{array}{lll} \mathbb{A}^2 - 0 & \to & \mathbb{A}^2 \\ & & \uparrow \\ & & * \end{array}$$ The horizontal morphism is just the inclusion. If you give me a $k$-homomorphism $\phi\colon k[x, y] \to k$ then that yields a vertical morphism because $\mathbb{A}^2$ is affine. I claim that if $\phi$ corresponded to a morphism $\pi\colon * \to \mathbb{A}^2 - 0$ then $\pi$ would make the above diagram into a commutative triangle.

[Note that in the bijection you give we only need that $W$ is affine.]

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  • $\begingroup$ I apologize for giving practically the same answer you did! Hopefully it added something. $\endgroup$ – Alex Youcis Nov 24 '14 at 8:21

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