**Proof.**
Part (1) follows from Lemma 89.16.12 and the fact that $\mathcal{W}(k)$ is the fibre product of two setoids with a unique isomorphism class over a setoid with a unique isomorphism class.

Part (2). Let $w_0 \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{W}(k))$ and let $x_0, y_0, z_0$ be the image of $w_0$ in $\mathcal{F}, \mathcal{H}, \mathcal{G}$. Then $\text{Inf}(\mathcal{W}) = \text{Inf}_{w_0}(\mathcal{W})$ and simlarly for $\mathcal{H}$, $\mathcal{G}$, and $\mathcal{F}$, see Remark 89.19.4. We apply Lemmas 89.12.4 and 89.19.10 to get all the linear maps except for the “boundary map” $\delta : \text{Inf}_{x_0}(\mathcal{F}) \to T\mathcal{W}$. We will insert suitable signs later.

Construction of $\delta $. Choose a pushforward $w_0 \to w'_0$ along $k \to k[\epsilon ]$. Denote $x'_0, y'_0, z'_0$ the images of $w'_0$ in $\mathcal{F}, \mathcal{H}, \mathcal{G}$. In particular we obtain isomorphisms $b' : f(y'_0) \to x'_0$ and $c' : x'_0 \to g(z'_0)$. Denote $b : f(y_0) \to x_0$ and $c : x_0 \to g(z_0)$ the pushforwards along $k[\epsilon ] \to k$. Observe that this means $w'_0 = (k[\epsilon ], y'_0, z'_0, c' \circ b')$ and $w_0 = (k, y_0, z_0, c \circ b)$ in terms of the explicit form of the fibre product of categories, see Remarks 89.5.2 (13). Given $\alpha : x'_0 \to x'_0$ we set $\delta (\alpha ) = (k[\epsilon ], y'_0, z'_0, c' \circ \alpha \circ b')$ which is indeed an object of $\mathcal{W}$ over $k[\epsilon ]$ and comes with a morphism $(k[\epsilon ], y'_0, z'_0, c' \circ \alpha \circ b') \to w_0$ over $k[\epsilon ] \to k$ as $\alpha $ pushes forward to the identity over $k$. More generally, for any $k$-vector space $V$ we can define a map

\[ \text{Lift}(\text{id}_{x_0}, k[V]) \longrightarrow \text{Lift}(w_0, k[V]) \]

using exactly the same formulae. This construction is functorial in the vector space $V$ (details omitted). Hence $\delta $ is $k$-linear by an application of Lemma 89.11.5.

Having constructed these maps it is straightforward to show the sequence is exact. Injectivity of the first map comes from the fact that $f \times g : \mathcal{W} \to \mathcal{H} \times \mathcal{G}$ is faithful. If $(\beta , \gamma ) \in \text{Inf}_{y_0}(\mathcal{H}) \oplus \text{Inf}_{z_0}(\mathcal{G})$ map to the same element of $\text{Inf}_{x_0}(\mathcal{F})$ then $(\beta , \gamma )$ defines an automorphism of $w'_0 = (k[\epsilon ], y'_0, z'_0, c' \circ b')$ whence exactness at the second spot. If $\alpha $ as above gives the trivial deformation $(k[\epsilon ], y'_0, z'_0, c' \circ \alpha \circ b')$ of $w_0$, then the isomorphism $w'_0 = (k[\epsilon ], y'_0, z'_0, c' \circ b') \to (k[\epsilon ], y'_0, z'_0, c' \circ \alpha \circ b')$ produces a pair $(\beta , \gamma )$ which is a preimage of $\alpha $. If $w = (k[\epsilon ], y, z, \phi )$ is a deformation of $w_0$ such that $y'_0 \cong y$ and $z \cong z'_0$ then the map

\[ f(y'_0) \to f(y) \xrightarrow {\phi } g(z) \to g(z'_0) \]

is an $\alpha $ which maps to $w$ under $\delta $. Finally, if $y$ and $z$ are deformations of $y_0$ and $z_0$ and there exists an isomorphism $\phi : f(y) \to g(z)$ of deformations of $f(y_0) = x_0 = g(z_0)$ then we get a preimage $w = (k[\epsilon ], y, z, \phi )$ of $(x, y)$ in $T\mathcal{W}$. This finishes the proof.
$\square$

## Comments (0)