The Stacks project

95.7 Surjective on objects

Let $S$ be a scheme. Let $p : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$. We will say that $p$ is surjective on objects if the following condition holds: Given any data consisting of

  1. a field $k$ over $S$, and

  2. an object $y$ of $\mathcal{Y}$ over $\mathop{\mathrm{Spec}}(k)$,

then there exists an extension $k \subset K$ of fields over $S$, an object $x$ of $\mathcal{X}$ over $\mathop{\mathrm{Spec}}(K)$ such that $p(x) \cong y|_{\mathop{\mathrm{Spec}}(K)}$.

Lemma 95.7.1. Let $p : \mathcal{X} \to \mathcal{Y}$ and $q : \mathcal{Z} \to \mathcal{Y}$ be $1$-morphisms of categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$. If $p : \mathcal{X} \to \mathcal{Y}$ is surjective on objects, then so is the base change $p' : \mathcal{X} \times _\mathcal {Y} \mathcal{Z} \to \mathcal{Z}$ of $p$ by $q$.

Proof. This is formal. Let $z$ be an object of $\mathcal{Z}$ over a field $k$. As $p$ is surjective on objects there exists an extension $k \subset K$ and an object $x$ of $\mathcal{X}$ over $K$ and an isomorphism $\alpha : p(x) \to q(z)|_{\mathop{\mathrm{Spec}}(K)}$. Then $w = (\mathop{\mathrm{Spec}}(K), x, z|_{\mathop{\mathrm{Spec}}(K)}, \alpha )$ is an object of $\mathcal{X} \times _\mathcal {Y} \mathcal{Z}$ over $K$ with $p'(w) = z|_{\mathop{\mathrm{Spec}}(K)}$. $\square$

Lemma 95.7.2. Let $p : \mathcal{X} \to \mathcal{Y}$ and $q : \mathcal{Y} \to \mathcal{Z}$ be $1$-morphisms of categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$. If $p$ and $q$ are surjective on objects, then so is the composition $q \circ p$.

Proof. This is formal. Let $z$ be an object of $\mathcal{Z}$ over a field $k$. As $q$ is surjective on objects there exists a field extension $k \subset K$ and an object $y$ of $\mathcal{Y}$ over $K$ such that $q(y) \cong x|_{\mathop{\mathrm{Spec}}(K)}$. As $p$ is surjective on objects there exists a field extension $K \subset L$ and an object $x$ of $\mathcal{X}$ over $L$ such that $p(x) \cong y|_{\mathop{\mathrm{Spec}}(L)}$. Then the field extension $k \subset L$ and the object $x$ of $\mathcal{X}$ over $L$ satisfy $q(p(x)) \cong z|_{\mathop{\mathrm{Spec}}(L)}$ as desired. $\square$

Lemma 95.7.3. Let $p : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of categories fibred in groupoids over $(\mathit{Sch}/S)_{fppf}$. If $p$ is representable by algebraic spaces, then the following are equivalent:

  1. $p$ is surjective on objects, and

  2. $p$ is surjective (see Algebraic Stacks, Definition 92.10.1).

Proof. Assume (2). Let $k$ be a field and let $y$ be an object of $\mathcal{Y}$ over $k$. Let $X_ y$ denote an algebraic space over $k$ representing the $2$-fibre product

\[ (\mathit{Sch}/\mathop{\mathrm{Spec}}(k))_{fppf} \times _{y, \mathcal{Y}, p} \mathcal{X}. \]

As we've assumed that $p$ is surjective we see that $X_ y$ is not empty. Hence we can find a field extension $k \subset K$ and a $K$-valued point $x$ of $X_ y$. Via the $2$-Yoneda lemma this corresponds to an object $x$ of $\mathcal{X}$ over $K$ together with an isomorphism $p(x) \cong y|_{\mathop{\mathrm{Spec}}(K)}$ and we see that (1) holds.

Assume (1). Choose a scheme $T$ and a $1$-morphism $y : (\mathit{Sch}/T)_{fppf} \to \mathcal{Y}$. Let $X_ y$ be an algebraic space over $T$ representing the $2$-fibre product $(\mathit{Sch}/T)_{fppf} \times _{y, \mathcal{Y}, p} \mathcal{X}$. We have to show that $X_ y \to T$ is surjective. By Morphisms of Spaces, Definition 65.5.2 we have to show that $|X_ y| \to |T|$ is surjective. This means exactly that given a field $k$ over $T$ and a morphism $t : \mathop{\mathrm{Spec}}(k) \to T$ there exists a field extension $k \subset K$ and a morphism $x : \mathop{\mathrm{Spec}}(K) \to X_ y$ such that

\[ \xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[d] \ar[r]_ x & X_ y \ar[d] \\ \mathop{\mathrm{Spec}}(k) \ar[r]^ t & T } \]

commutes. By the $2$-Yoneda lemma this means exactly that we have to find $k \subset K$ and an object $x$ of $\mathcal{X}$ over $K$ such that $p(x) \cong t^*y|_{\mathop{\mathrm{Spec}}(K)}$. Hence (1) guarantees that this is the case and we win. $\square$


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