Lemma 93.10.3. In Example 93.9.1 let $f : X \to Y$ be a morphism of schemes over $k$. There is a canonical exact sequence of $k$-vector spaces
\[ \xymatrix{ 0 \ar[r] & \text{Inf}(\mathcal{D}\! \mathit{ef}_{X \to Y}) \ar[r] & \text{Inf}(\mathcal{D}\! \mathit{ef}_ X \times \mathcal{D}\! \mathit{ef}_ Y) \ar[r] & \text{Der}_ k(\mathcal{O}_ Y, f_*\mathcal{O}_ X) \ar[lld] \\ & T\mathcal{D}\! \mathit{ef}_{X \to Y} \ar[r] & T(\mathcal{D}\! \mathit{ef}_ X \times \mathcal{D}\! \mathit{ef}_ Y) \ar[r] & \text{Ext}^1_{\mathcal{O}_ X}(Lf^*\mathop{N\! L}\nolimits _{Y/k}, \mathcal{O}_ X) } \]
Proof.
The obvious map of deformation categories $\mathcal{D}\! \mathit{ef}_{X \to Y} \to \mathcal{D}\! \mathit{ef}_ X \times \mathcal{D}\! \mathit{ef}_ Y$ gives two of the arrows in the exact sequence of the lemma. Recall that $\text{Inf}(\mathcal{D}\! \mathit{ef}_{X \to Y})$ is the set of automorphisms of the trivial deformation
\[ f' : X' = X \times _{\mathop{\mathrm{Spec}}(k)} \mathop{\mathrm{Spec}}(k[\epsilon ]) \xrightarrow {f \times \text{id}} Y' = Y \times _{\mathop{\mathrm{Spec}}(k)} \mathop{\mathrm{Spec}}(k[\epsilon ]) \]
of $X \to Y$ to $k[\epsilon ]$ equal to the identity modulo $\epsilon $. This is clearly the same thing as pairs $(\alpha , \beta ) \in \text{Inf}(\mathcal{D}\! \mathit{ef}_ X \times \mathcal{D}\! \mathit{ef}_ Y)$ of infinitesimal automorphisms of $X$ and $Y$ compatible with $f'$, i.e., such that $f' \circ \alpha = \beta \circ f'$. By Deformation Theory, Lemma 91.7.1 for an arbitrary pair $(\alpha , \beta )$ the difference between the morphism $f' : X' \to Y'$ and the morphism $\beta ^{-1} \circ f' \circ \alpha : X' \to Y'$ defines an element in
\[ \text{Der}_ k(\mathcal{O}_ Y, f_*\mathcal{O}_ X) = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_ Y}(\Omega _{Y/k}, f_*\mathcal{O}_ X) \]
Equality by More on Morphisms, Lemma 37.13.3. This defines the last top horizontal arrow and shows exactness in the first two places. For the map
\[ \text{Der}_ k(\mathcal{O}_ Y, f_*\mathcal{O}_ X) \to T\mathcal{D}\! \mathit{ef}_{X \to Y} \]
we interpret elements of the source as morphisms $f_\epsilon : X' \to Y'$ over $\mathop{\mathrm{Spec}}(k[\epsilon ])$ equal to $f$ modulo $\epsilon $ using Deformation Theory, Lemma 91.7.1. We send $f_\epsilon $ to the isomorphism class of $(f_\epsilon : X' \to Y')$ in $T\mathcal{D}\! \mathit{ef}_{X \to Y}$. Note that $(f_\epsilon : X' \to Y')$ is isomorphic to the trivial deformation $(f' : X' \to Y')$ exactly when $f_\epsilon = \beta ^{-1} \circ f \circ \alpha $ for some pair $(\alpha , \beta )$ which implies exactness in the third spot. Clearly, if some first order deformation $(f_\epsilon : X_\epsilon \to Y_\epsilon )$ maps to zero in $T(\mathcal{D}\! \mathit{ef}_ X \times \mathcal{D}\! \mathit{ef}_ Y)$, then we can choose isomorphisms $X' \to X_\epsilon $ and $Y' \to Y_\epsilon $ and we conclude we are in the image of the south-west arrow. Therefore we have exactness at the fourth spot. Finally, given two first order deformations $X_\epsilon $, $Y_\epsilon $ of $X$, $Y$ there is an obstruction in
\[ ob(X_\epsilon , Y_\epsilon ) \in \text{Ext}^1_{\mathcal{O}_ X}(Lf^*\mathop{N\! L}\nolimits _{Y/k}, \mathcal{O}_ X) \]
which vanishes if and only if $f : X \to Y$ lifts to $X_\epsilon \to Y_\epsilon $, see Deformation Theory, Lemma 91.7.1. This finishes the proof.
$\square$
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