## 95.6 Formally smooth 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 formally smooth on objects if the following condition holds: Given any data consisting of

1. a first order thickening $U \subset U'$ of affine schemes over $S$,

2. an object $y'$ of $\mathcal{Y}$ over $U'$,

3. an object $x$ of $\mathcal{X}$ over $U$, and

4. an isomorphism $\gamma : p(x) \to y'|_ U$,

then there exists an object $x'$ of $\mathcal{X}$ over $U'$ with an isomorphism $\beta : x'|_ U \to x$ and an isomorphism $\gamma ' : p(x') \to y'$ such that

95.6.0.1
$$\label{criteria-equation-formally-smooth} \vcenter { \xymatrix{ p(x'|_ U) \ar[d]_{p(\beta )} \ar[rr]_{\gamma '|_ U} & & y'|_ U \ar@{=}[d] \\ p(x) \ar[rr]^\gamma & & y'|_ U } }$$

commutes. In this situation we say that “$(x', \beta , \gamma ')$ is a solution to the problem posed by our data (1), (2), (3), (4)”.

Lemma 95.6.1. 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 formally smooth 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 $U \subset U'$ be a first order thickening of affine schemes over $S$, let $z'$ be an object of $\mathcal{Z}$ over $U'$, let $w$ be an object of $\mathcal{X} \times _\mathcal {Y} \mathcal{Z}$ over $U$, and let $\delta : p'(w) \to z'|_ U$ be an isomorphism. We may write $w = (U, x, z, \alpha )$ for some object $x$ of $\mathcal{X}$ over $U$ and object $z$ of $\mathcal{Z}$ over $U$ and isomorphism $\alpha : p(x) \to q(z)$. Note that $p'(w) = z$ hence $\delta : z \to z|_ U$. Set $y' = q(z')$ and $\gamma = q(\delta ) \circ \alpha : p(x) \to y'|_ U$. As $p$ is formally smooth on objects there exists an object $x'$ of $\mathcal{X}$ over $U'$ as well as isomorphisms $\beta : x'|_ U \to x$ and $\gamma ' : p(x') \to y'$ such that (95.6.0.1) commutes. Then we consider the object $w = (U', x', z', \gamma ')$ of $\mathcal{X} \times _\mathcal {Y} \mathcal{Z}$ over $U'$ and define isomorphisms

$w'|_ U = (U, x'|_ U, z'|_ U, \gamma '|_ U) \xrightarrow {(\beta , \delta ^{-1})} (U, x, z, \alpha ) = w$

and

$p'(w') = z' \xrightarrow {\text{id}} z'.$

These combine to give a solution to the problem. $\square$

Lemma 95.6.2. 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 formally smooth on objects, then so is the composition $q \circ p$.

Proof. This is formal. Let $U \subset U'$ be a first order thickening of affine schemes over $S$, let $z'$ be an object of $\mathcal{Z}$ over $U'$, let $x$ be an object of $\mathcal{X}$ over $U$, and let $\gamma : q(p(x)) \to z'|_ U$ be an isomorphism. As $q$ is formally smooth on objects there exist an object $y'$ of $\mathcal{Y}$ over $U'$, an isomorphism $\beta : y'|_ U \to p(x)$, and an isomorphism $\gamma ' : q(y') \to z'$ such that (95.6.0.1) is commutative. As $p$ is formally smooth on objects there exist an object $x'$ of $\mathcal{X}$ over $U'$, an isomorphism $\beta ' : x'|_ U \to x$, and an isomorphism $\gamma '' : p(x') \to y'$ such that (95.6.0.1) is commutative. The solution is to take $x'$ over $U'$ with isomorphism

$q(p(x')) \xrightarrow {q(\gamma '')} q(y') \xrightarrow {\gamma '} z'$

and isomorphism $\beta ' : x'|_ U \to x$. We omit the verification that (95.6.0.1) is commutative. $\square$

Note that the class of formally smooth morphisms of algebraic spaces is stable under arbitrary base change and local on the target in the fpqc topology, see More on Morphisms of Spaces, Lemma 74.19.3 and 74.19.11. Hence condition (2) in the lemma below makes sense.

Lemma 95.6.3. 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 formally smooth on objects, and

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

Proof. Assume (2). Let $U \subset U'$ be a first order thickening of affine schemes over $S$, let $y'$ be an object of $\mathcal{Y}$ over $U'$, let $x$ be an object of $\mathcal{X}$ over $U$, and let $\gamma : p(x) \to y'|_ U$ be an isomorphism. Let $X_{y'}$ denote an algebraic space over $U'$ representing the $2$-fibre product

$(\mathit{Sch}/U')_{fppf} \times _{y', \mathcal{Y}, p} \mathcal{X}.$

Note that $\xi = (U, U \to U', x, \gamma ^{-1})$ defines an object of this $2$-fibre product over $U$. Via the $2$-Yoneda lemma $\xi$ corresponds to a morphism $f_\xi : U \to X_{y'}$ over $U'$. As $X_{y'} \to U'$ is formally smooth by assumption there exists a morphism $f' : U' \to X_{y'}$ such that $f_\xi$ is the composition of $f'$ and the morphism $U \to U'$. Also, the $2$-Yoneda lemma tells us that $f'$ corresponds to an object $\xi ' = (U', U' \to U', x', \alpha )$ of the displayed $2$-fibre product over $U'$ whose restriction to $U$ recovers $\xi$. In particular we obtain an isomorphism $\gamma : x'|U \to x$. Note that $\alpha : y' \to p(x')$. Hence we see that taking $x'$, the isomorphism $\gamma : x'|U \to x$, and the isomorphism $\beta = \alpha ^{-1} : p(x') \to y'$ is a solution to the problem.

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 formally smooth. Hence it suffices to show that given a first order thickening $U \subset U'$ of affine schemes over $T$, then $X_ y(U') \to X_ y(U')$ is surjective (morphisms in the category of algebraic spaces over $T$). Set $y' = y|_{U'}$. By the $2$-Yoneda lemma morphisms $U \to X_ y$ over $T$ correspond bijectively to isomorphism classes of pairs $(x, \alpha )$ where $x$ is an object of $\mathcal{X}$ over $U$ and $\alpha : y|_ U \to p(x)$ is an isomorphism. Of course giving $\alpha$ is, up to an inverse, the same thing as giving an isomorphism $\gamma : p(x) \to y'|_ U$. Similarly for morphisms $U' \to X_ y$ over $T$. Hence (1) guarantees the surjectivity of $X_ y(U') \to X_ y(U')$ in this situation and we win. $\square$

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