I have a smooth map $\Gamma : X\times Y \to Z$ between manifolds with or without boundary, and I am interested in the maps $\Gamma_y : X \to Z$ given by $\Gamma_y(x) = \Gamma(x,y)$. For a given $y\in Y$, how are the differentials $d(\Gamma_y)_x : T_xX \to T_{\Gamma_y(x)}Z$ and $d\Gamma_{(x,y)} : T_{(x,y)}(X\times Y) \to T_{\Gamma(x,y)}Z$ related?
I know at least that $T_{(x,y)}(X\times Y) \cong T_xX \oplus T_yY$, so $d(\Gamma_y)_x$ and $d\Gamma_{(x,y)}$ definitely cannot be the same map. But, for example, is there a tangent vector $\mathbf{v}_y \in T_yY$ such that $$ d\Gamma_{(x,y)}(\mathbf{u}_x, \mathbf{v}_y) = d(\Gamma_y)_x(\mathbf{u}_x) $$ for all $\mathbf{u}_x \in T_xX$? If so, what is this $\mathbf{v}_y$?
$\Gamma_y$ is the composition $$X \stackrel{i_y}\longrightarrow X \times Y \stackrel{\Gamma}\longrightarrow Z$$ where the first map is $i_y(x) = (x,y)$.
Then, with the identification $T_{(x,y)}(X \times Y) = T_{x}X \oplus T_yY$ we see that $d(i_y)_x(v_x) = (v_x, 0_y)$. Hence, $$d(\Gamma_y)_x(v_x) = d(\Gamma \circ i_y)_x(v_x) = d\Gamma_{(x,y)}((v_x, 0_y)).$$