Let $X$ be a proper scheme over a field $k$. Let $T$ be a scheme over $k$. Is it true that morphisms $T \times X \to \mathbb{A}^1$ are in bijection with morphisms $T \to \Gamma (X, \mathcal{O}\_X)$ (where the finite-dimensional vector space $\Gamma (X, \mathcal{O}\_X)$ is interpreted as a scheme as usual)?


Or, perhaps, is something weaker (put condition on $T$, say) or stronger (more general scheme instead of $\mathbb{A}^1$, say) true?


Thank you



Mapping scheme from a proper variety
You can just use all the universal properties one at a time. We only need that $X \to \operatorname{Spec} k$ is qcqs and that $\Gamma(X,\mathcal O\_X)$ is finite-dimensional; both these assumptions are satisfied if $X$ is proper over $k$. We also need that $T \to \operatorname{Spec} k$ is flat, which is always true if $k$ is a field.


**Remark.** Note that $\Gamma(T \times X, \mathcal O\_{T \times X}) = \Gamma(T, \mathcal O\_T) \otimes \Gamma(X, \mathcal O\_X)$. Indeed, this follows from flat base change ([Tag 02KH](http://stacks.math.columbia.edu/tag/02KH)), since $T \to \operatorname{Spec} k$ is flat and $X \to \operatorname{Spec} k$ is qcqs.


**Remark.** If $V$ is a vector space and $W$ is a finite-dimensional vector space, then the natural map
\begin{align\*}
V \otimes W &\to \operatorname{Hom}\_k(W^\vee, V)\\
v \otimes w &\mapsto (\phi \mapsto \phi(w) v)
\end{align\*}
is a natural isomorphism. For example, one can prove the case for $W$ of dimension $1$, and use that every $W$ decomposes as a finite direct sum of $1$-dimensional vector spaces. (Note, however, that this is false for $W$ infinite-dimensional, already if $V = k$.)



> 
> **Lemma.** Let $X$ be a qcqs $k$-scheme such that $\Gamma(X, \mathcal O\_X)$ is finite-dimensional. Let $T$ be a $k$-scheme. Then
> $$\operatorname{Hom}\_{\textrm{Sch}/k}(T \times X, \mathbb A^1) = \operatorname{Hom}\_{\textrm{Sch}/k}(T, \operatorname{Spec} \operatorname{S}(\Gamma(X,\mathcal O\_X)^\vee)).$$
> 


*Proof.* By the adjunction $\operatorname{Ring}^{\operatorname{op}} \rightleftarrows \operatorname{Sch}$ given by $\Gamma$ and $\operatorname{Spec}$, we get
$$\operatorname{Hom}\_{\textrm{Sch}/k}(T \times X, \mathbb A^1) = \operatorname{Hom}\_k^{\operatorname{alg}}(k[x],\Gamma(T \times X, \mathcal O\_{T \times X})).$$
By the universal property of $k[x]$, the latter is just $\Gamma(T \times X, \mathcal O\_{T \times X})$. By the two remarks above, this equals
$$\Gamma(T, \mathcal O\_T) \otimes \Gamma(X, \mathcal O\_X) = \operatorname{Hom}\_k(\Gamma(X,\mathcal O\_X)^\vee, \Gamma(T,\mathcal O\_T)).$$
The universal property of the symmetric algebra turns this into
$$\operatorname{Hom}\_k^{\operatorname{alg}}(\operatorname{S}(\Gamma(X, \mathcal O\_X)^\vee), \Gamma(T,\mathcal O\_T)).$$
Finally, using the adjunction $\operatorname{Ring}^{\operatorname{op}} \rightleftarrows \operatorname{Sch}$ again, this finally becomes
$$\operatorname{Hom}\_{\textrm{Sch}/k}(T, \operatorname{Spec} \operatorname{S}(\Gamma(X,\mathcal O\_X)^\vee)),$$
which proves the claim. $\square$


