> [!theorem] > > Let $E$ be a [[Banach Space|Banach space]], $X$ be an $E$-[[Manifold|manifold]], and $p \in X$. Let $T_pX$ be the [[Tangent Space|tangent space]] at $p$, then $T_pX$ is [[Space of Toplinear Isomorphisms|toplinearly isomorphic]] to $T$. > > *Proof.* > > *Warning: informal argument.* > > Our generating set for [$\widehat{\topo}$](Topological%Space) consists of pairwise disjoint open sets constrained to each [[Concrete Tangent Space|concrete space]]. Therefore $\widehat{\topo}$ consists *only* of unions of these sets. > > Designate one $i \in I$ as the reference space, and $T_j = D(\psi_{i} \circ \psi_j^{-1})(p)$ be the isomorphism from $T_{(U_j, \psi_j, p)}$ to $T_{(U_i, \psi_i, p)}$. Then for any $U_j \subset T_{(U_j, \psi_j, p)}$ open, $T_j(U_j)$ is open in $T_{(U_i, \psi_i, p)}$, so $U_j$ and $T_j(U_j)$ have equivalent elements. As every open set is a union of sets of the form $U_j$, projecting all of them onto $T_{(U_i, \psi_i, p)}$ yields the equivalent open set. Therefore every open set in the [[Quotient Topology|quotient topology]] corresponds to an open set on $T_{(U_i, \psi_i, p)}$. As $T_{(U_i, \psi_i, p)}$ is toplinearly isomorphic to $E$, so is the tangent space.