Part (1) follows from Lemma \ref{lemma-RS-fiber-product-morphisms} 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. \medskip\noindent Part (2). Let $w_0 \in \Ob(\mathcal{W}(k))$ 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 \ref{remark-trivial-aut-point}. We apply Lemmas \ref{lemma-k-linear-differential} and \ref{lemma-k-linear-infaut} 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. \medskip\noindent 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 \ref{remarks-cofibered-groupoids} (\ref{item-fibre-product}). 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 \ref{lemma-morphism-linear-functors}. \medskip\noindent 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.