# Can Yoneda lemma for smooth projective varieties only use curves?

Let $$X,Y$$ be two projective smooth varieties over an algebraically closed field $$k$$. If we have natural isomorphism $$\operatorname{Hom}(C,X) \cong \operatorname{Hom}(C,Y)$$ as sets for every smooth projective curve $$C$$, then do we know $$X \cong Y$$?

In other words, $$Hom(-,Y)$$ and $$Hom(-,X)$$ are two isomorphic functors when restricted to the subcategory of curves. If not, for a fixed $$X=X_0$$, can we classify all such $$Y$$ ? Are they all birational? When is there a unique $$Y$$ (namely $$X_0$$)?

What if we require all curves $$C$$ (not necessarily smooth or projective)?

• To avoid confusion, you should use the phrase "natural isomorphism" rather than "functorial isomorphism". I assume that what you mean is that there is a natural isomorphism between the contravariant functors $\text{Hom}(-,X)$ and $\text{Hom}(-,Y)$ from the category of smooth projective curves to the category of sets, right? Dec 30, 2019 at 21:19

This is a twice updated version:

• Thanks to @PiotrPstrągowski for pointing out some issues that require more care.
• Some thoughts were in response to the original wording of the question. I left them here, because they might be interesting to some.
• As @AndréHenriques points out this still doesn't work -- at least compared to the updated question. My initial reaction was to delete the answer, but perhaps it is still interesting, so I will leave it here.

The main issue here is to decide what you mean by "functorial isomorphism"? If, for instance, you mean that it is induced by a morphism $$\phi:X\to Y$$, then sure, it is true: By your condition $$\phi$$ would have to be surjective and have only zero dimensional fibers, so by ZMT an isomorphism.

On the other hand, if we're not asking for such a strong funtoriality, only for some "natural" way to have the two sets be bijective, then one can do this.

As @AndréHenriques points out this does not work if we allow constant maps between curves. For whatever it is worth it still seems to work if we restrict to dominant maps...

Choose an $$n\in\mathbb N$$, $$n\gg 0$$ and choose two disjoint sets of $$n$$-points in general position in $$\mathbb P^2$$, say $$p_i$$ and $$q_i$$, such that there is no automorphism of $$\mathbb P^2$$ that takes one set to the other. Let $$X$$ and $$Y$$ be $$\mathbb P^2$$ blown-up along these two sets of points respectively (say $$X$$ along the $$p_i$$'s and $$Y$$ along the $$q_i$$'s).

For any $$C\to X$$ or $$C\to Y$$ that when composed with the blow-down to $$\mathbb P^2$$, it does not map into any $$p_i$$'s or $$q_i$$'s, there is a well-defined proper transform on the other one (between $$X$$ and $$Y$$), so make these correspond to each other.

For each $$p_i$$ and $$q_i$$ let the corresponding exceptional curves be $$E_i\subset X$$ and $$F_i\subset Y$$ respectively and choose a point on each: $$P_i\in E_i$$ and $$Q_i\in F_i$$. In addition, choose an isomorphism $$\phi_i:E_i\to F_i$$ such that $$\phi_i(P_i)=Q_i$$.

Now let $$\alpha:C\to E_i\subset X$$ be a morphism. If $$\alpha(C)=\{P_i\}$$ then make it correspond to the "same" map $$\alpha':C\to p_i\in Y$$, otherwise make it correspond to the map $$\alpha''=\phi_i\circ\alpha:C\to F_i\subset Y$$. Finally, let $$\alpha:C\to q_i\in X$$ be a morphism and make it correspond to the "same" map $$\alpha':C\to _i\in Y$$. This, with the proper transforms above, gives a bijection $$\operatorname{Hom}(C,X)\leftrightarrow \operatorname{Hom}(C,Y)$$ which is functorial in $$C$$.

Note that this is for the case when $$C$$ is smooth or at least irreducible. If we allow singular reducible curves, then this does not work, because one could consider maps from two intersecting lines, one mapping to an exceptional curve, say $$E_i$$, and the other to the proper transform on $$X$$ of a general line through $$P_i\in\mathbb P^2$$. If the bijection between the $$\operatorname{Hom}$$'s is functorial, then it would have to map the proper transform to its proper transform on $$Y$$ which will not intersect the image of the other line by the morphism that is assigned to it by the above process.

As @PiotrPstrągowski points out, the assumptions imply that there is a bijection between the points of $$X$$ and the points of $$Y$$. The above example shows that as long as $$C$$ is smooth, this bijection does not have to be a morphism, and accordingly $$X$$ and $$Y$$ are not necessarily isomorphic. However, it does leave the possibility open that

• $$X$$ and $$Y$$ may be necessarily birational, and
• Allowing $$C$$ to be singular may force $$X$$ and $$Y$$ to be isomorphic.

Remark: I don't think allowing non-projective curves would make a difference as long as $$X$$ and $$Y$$ remain projective.

• I mean the isomorphism is functorial with respect to the curve $C$ (not assumed to be induced by a morphism). Thank you for the example, so do we know that $X$ and $Y$ must be birational? Dec 30, 2019 at 20:44
• OK. It seems to me that the example I gave is functorial in $C$. It does not work if you allow singular curves for $C$, but I would expect that there is a more complicated example for that case. It occurred to me as well, that the next natural question is whether $X$ and $Y$ would have to be birational. At the moment, I don't know. Dec 31, 2019 at 3:43
• So, I don't think allowing all kinds of curves helps... Dec 31, 2019 at 3:50
• Thank you! These examples are interesting. Dec 31, 2019 at 6:04
• @SándorKovács Since $Aut(\mathbb{P}^{1})$ exchanges any pair of two points, I believe that we can identify the underlying set of a variety X with those points in $Hom(\mathbb{P}^{1}, X)$ which are fixed points for the automorphism group of $\mathbb{P}^{1}$. It follows that whenever we have an isomorphism $Hom(C, X) \simeq Hom(C, Y)$ natural in $C$, then we get a bijection $\phi: X \rightarrow Y$. (Moreover, this $\phi$ will have the property that for any $f: C \rightarrow X$ from a smooth proj. curve, $f$ is regular if and only if $\phi \circ f$ is.) What would this $\phi$ be in your example? Dec 31, 2019 at 19:27