Lang's text discusses differentiable functors of two variables, but that appears to be very much out of reach for now, so I will reduce this to the case of one variables. > [!definition] > > Let $\mathfrak{A}$ and $\mathfrak{B}$ be [[Category|subcategories]] of [[Banach Space|Banach spaces]], and $\lambda: \cata \to \mathfrak{B}$ be a contravariant [[Functor|functor]]. By definition for any $E, F \in \mathfrak{A}$, this induces a map > $ > \lambda: L(E, F) \to L(\lambda(F), \lambda(E)) > $ > If for any [[Manifold|manifold]] $U$ and $C^p$ morphism $\varphi: U \to L(E, F)$, the composition > $ > \begin{CD} > U @>{\varphi}>> L(E, F) @>{\lambda}>> L(\lambda(F), \lambda(E)) > \end{CD} > $ > is [$C^p$](Space%20of%20Continuously%20Differentiable%20Functions), then $\lambda$ is a **class** $C^p$. > > Remark: while the map between $L(E, F) \to L(\lambda(F), \lambda(E))$ being $C^p$ would satisfy the desired condition, this does not seem necessary. > [!theoremb] Theorem > > Let $\mathfrak{A}, \mathfrak{B}$ be subcategories of Banach spaces, $\lambda: \mathfrak{A} \to \mathfrak{B}$ be a functor of class $C^p$, and $X$ be a $C^{p + 1}$ [[Manifold|manifold]] modelled on $E$. Then $\lambda$ induces a functor between [[Vector Bundle|vector bundles]] over $X$ > $ > \lambda_X: \text{VB}(X, \mathfrak{A}) \to \text{VB}(X, \mathfrak{B}) > $ > such that for any bundle $\alpha, \beta \in \text{VB}(X, \mathfrak{A})$, and [[VB-Morphism|VB-morphism]] $f: \alpha \to \beta$, > 1. $\lambda_X(\alpha)_x = \lambda(\alpha_x)$ > 2. $\lambda_X(f)_x = \lambda(f_x)$ > 3. If $\alpha$ is the trivial bundle $X \times F$, then $\lambda_X(\alpha)$ is the trivial bundle $X \times \lambda(F)$. > > *Proof*. Let $\alpha \in \text{VB}(X, \mathfrak{A})$, then for each $p \in X$, $\alpha_p \in \cata$, which allows defining > $ > \lambda_X(\alpha) = \bigsqcup_{p \in X}\lambda(\alpha_p) > $ > which automatically makes $\lambda_X(\alpha)$ satisfy $(1)$. Let > $ > \tau: \alpha^{-1}(U) \to U \times F \quad \tau_p \in L(\alpha_p, F) > $ > be a trivialising map on $U$. Define > $ > \rho: \lambda_X(\alpha)^{-1}(U) \to U \times \lambda(F) \quad \rho_p = \lambda(\tau_p^{-1}) \in L(\lambda(\alpha_p), \lambda(F)) > $ > then $\rho$ is a trivialising map on $\lambda_X(\alpha)$. To verify the compatibility between charts, we find that for any trivialising maps $\tau_1$ and $\tau_2$ with overlapping maps, the corresponding maps $\rho_1$ and $\rho_2$ satisfy > $ > \begin{align*} > (\rho_1 \circ \rho_2^{-1})_p &= \lambda({\tau_1}_{p}^{-1}) \circ \lambda({\tau_2}_{{p}}^{-1})^{-1} \\ > &= \lambda({\tau_1}_{p}^{-1}) \circ \lambda({\tau_2}_{p}) \\ > &= \lambda({\tau_2}_{p} \circ {\tau_1}_{p}^{-1}) \\ > &= \lambda\braks{(\tau_{2} \circ \tau_1^{-1})_p} > \end{align*} > $ > which can be expressed as a composition of two $C^p$ maps, and hence is also $C^p$. > > Now let $\alpha, \beta \in \text{VB}(X, \mathfrak{A})$ be modelled on $F_1$ and $F_2$ respectively, and $f: \alpha \to \beta$ be a [[VB-Morphism|VB-morphism]]. Let $f_M: X \to X$ be the manifold part of the morphism. Define $\lambda_X(f): \beta \to \alpha$ with $\lambda_X(f)_M = f_m$, and on each fibre, define > $ > \lambda_X(f)_x = \lambda(f_x) \in L(\lambda(\beta_p), \lambda(\alpha_p)) > $ > then $\lambda_X(f)$ satisfies the projection compatibility condition. > > Since $f$ is a VB-morphism, for any $p \in X$, there exists trivialising maps $\tau_1$ for $\alpha$ and $\tau_2$ for $\beta$ with domain on $\alpha^{-1}(U)$ and $\beta^{-1}(U)$ respectively, such that the map > $ > U \to L(F_1, F_2) \quad x \mapsto (\tau_2)_{f_M(x)} \circ (f_B)_x \circ (\tau_1^{-1})_{\tau_1(x)} > $ > is $C^p$. With the corresponding maps $\rho_1$ and $\rho_2$, we get > $ > U \to L(\lambda(F_2), \lambda(F_1)) \quad x \mapsto (\rho_1)_{f_M(x)} \circ \lambda_X(f)_x \circ (\rho_2^{-1}) > $ > However since > $ > \begin{align*} > (\rho_1)_{f_M(x)} \circ \lambda_X(f)_x \circ (\rho_2^{-1}) &= \lambda\paren{{\tau_1}_{f_M(x)}^{-1}} \circ \lambda(f_x) \circ \lambda\paren{\tau_2} \\ > &= \lambda\braks{(\tau_2)_{f_M(x)} \circ (f_B)_x \circ (\tau_1^{-1})_{\tau_1(x)}} > \end{align*} > $ > and $\lambda$ is differentiable, $\lambda_X(f)$ satisfies the chart compatibility condition. > > Hence $\lambda_X$ is the desired functor.