Lemma 77.20.3. Assumptions and notation as in Lemma 77.20.2. There exists a canonical $2$-morphism $\alpha : \pi \circ s \to \pi \circ t$ making the diagram

$2$-commutative.

Lemma 77.20.3. Assumptions and notation as in Lemma 77.20.2. There exists a canonical $2$-morphism $\alpha : \pi \circ s \to \pi \circ t$ making the diagram

\[ \xymatrix{ \mathcal{S}_ R \ar[r]_ s \ar[d]_ t & \mathcal{S}_ U \ar[d]^\pi \\ \mathcal{S}_ U \ar[r]^-\pi & [U/R] } \]

$2$-commutative.

**Proof.**
Let $S'$ be a scheme over $S$. Let $r : S' \to R$ be a morphism over $S$. Then $r \in R(S')$ is an isomorphism between the objects $s \circ r, t \circ r \in U(S')$. Moreover, this construction is compatible with pullbacks. This gives a canonical $2$-morphism $\alpha _ p : \pi _ p \circ s \to \pi _ p \circ t$ where $\pi _ p : \mathcal{S}_ U \to [U/_{\! p}R]$ is as in the proof of Lemma 77.20.2. Thus even the diagram

\[ \xymatrix{ \mathcal{S}_ R \ar[r]_ s \ar[d]^ t & \mathcal{S}_ U \ar[d]^{\pi _ p} \\ \mathcal{S}_ U \ar[r]^-{\pi _ p} & [U/_{\! p}R] } \]

is $2$-commutative. Thus a fortiori the diagram of the lemma is $2$-commutative. $\square$

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