[proofplan] We use every smooth chart on $M$ to build a chart on $TM$ by recording the base point in local coordinates and the tangent vector in the corresponding coordinate basis. The only substantive compatibility check is the transition formula on overlaps: the fiber coordinate transforms by the Jacobian of the base transition map. Once the charts are compatible, we define the topology and smooth structure from this atlas, verify the manifold axioms, and observe that the projection becomes the coordinate projection $(x,v)\mapsto x$. [/proofplan]

[step:Construct tangent bundle charts from smooth coordinate charts] Let $(U,\varphi)$ be a smooth chart on $M$, so \begin{align*} \varphi: U &\to \varphi(U)\subseteq \mathbb{R}^n \end{align*} is a homeomorphism onto an open subset and has smooth transition maps with every compatible chart. Define \begin{align*} \widetilde{\varphi}: \pi^{-1}(U) &\to \varphi(U)\times \mathbb{R}^n \\ v_p &\mapsto \bigl(\varphi(p), d\varphi_p(v_p)\bigr). \end{align*} Here $v_p\in T_pM$, and $d\varphi_p:T_pM\to T_{\varphi(p)}\mathbb{R}^n\cong \mathbb{R}^n$ is the differential of $\varphi$ at $p$, using the standard identification of $T_{\varphi(p)}\mathbb{R}^n$ with $\mathbb{R}^n$ by constant coordinate vector fields. For each $p\in U$, the map $d\varphi_p:T_pM\to \mathbb{R}^n$ is a linear isomorphism, because $\varphi$ is a local diffeomorphism in coordinates. Therefore $\widetilde{\varphi}$ is bijective. Its inverse is the map \begin{align*} \widetilde{\varphi}^{-1}: \varphi(U)\times \mathbb{R}^n &\to \pi^{-1}(U) \\ (x,a) &\mapsto d(\varphi^{-1})_x(a), \end{align*} where $x\in \varphi(U)$, $a\in \mathbb{R}^n$, and $d(\varphi^{-1})_x:\mathbb{R}^n\to T_{\varphi^{-1}(x)}M$ is the differential of the coordinate inverse at $x$. Since $\varphi(U)$ is open in $\mathbb{R}^n$, the product $\varphi(U)\times \mathbb{R}^n$ is open in $\mathbb{R}^{2n}$. Thus each $\widetilde{\varphi}$ is a candidate coordinate chart of dimension $2n$. [/step]

[step:Compute the transition maps on tangent chart overlaps]Let $(U,\varphi)$ and $(V,\psi)$ be smooth charts on $M$ with $U\cap V\neq\varnothing$. Define the base transition map \begin{align*} F: \varphi(U\cap V) &\to \psi(U\cap V) \\ x &\mapsto \psi(\varphi^{-1}(x)). \end{align*} Since $M$ is a smooth manifold, $F=\psi\circ\varphi^{-1}$ is smooth. The tangent chart transition map is \begin{align*} \widetilde{\psi}\circ \widetilde{\varphi}^{-1}: \varphi(U\cap V)\times \mathbb{R}^n &\to \psi(U\cap V)\times \mathbb{R}^n. \end{align*} For $(x,a)\in \varphi(U\cap V)\times \mathbb{R}^n$, set $p=\varphi^{-1}(x)$ and $v_p=d(\varphi^{-1})_x(a)\in T_pM$. Then \begin{align*} (\widetilde{\psi}\circ \widetilde{\varphi}^{-1})(x,a) &= \widetilde{\psi}(v_p) \\ &= \bigl(\psi(p), d\psi_p(v_p)\bigr) \\ &= \bigl(\psi(\varphi^{-1}(x)), d\psi_{\varphi^{-1}(x)}(d(\varphi^{-1})_x(a))\bigr) \\ &= \bigl(F(x), dF_x(a)\bigr). \end{align*} Under the standard bases of $\mathbb{R}^n$, the [linear map](/page/Linear%20Map) $dF_x:\mathbb{R}^n\to\mathbb{R}^n$ is represented by the Jacobian matrix $JF_x$. Hence the transition map has coordinate formula \begin{align*} (x,a)\mapsto \bigl(F(x), JF_x\,a\bigr). \end{align*} Because $F$ is smooth, each component $F_i$ is smooth, and each partial derivative $\partial_{x_j}F_i$ is smooth. Therefore each component of $JF_x\,a$ is \begin{align*} \sum_{j=1}^n \partial_{x_j}F_i(x)\,a_j, \end{align*} which is smooth as a function of $(x,a)$. Thus $\widetilde{\psi}\circ\widetilde{\varphi}^{-1}$ is smooth. To check the inverse transition map, define \begin{align*} G: \psi(U\cap V) &\to \varphi(U\cap V) \\ y &\mapsto \varphi(\psi^{-1}(y)). \end{align*} Since $G=\varphi\circ\psi^{-1}$ is a smooth base transition map, the chain-rule computation above gives \begin{align*} (\widetilde{\varphi}\circ\widetilde{\psi}^{-1})(y,b)=\bigl(G(y),dG_y(b)\bigr)=\bigl(G(y),JG_y\,b\bigr). \end{align*} Each component of $JG_y\,b$ has the form \begin{align*} \sum_{j=1}^n \partial_{y_j}G_i(y)\,b_j, \end{align*} which is smooth as a function of $(y,b)$. Hence $\widetilde{\varphi}\circ\widetilde{\psi}^{-1}$ is smooth.[/step]

[guided]We must check that the proposed tangent charts are compatible. The overlap of the two tangent chart domains is \begin{align*} \pi^{-1}(U)\cap \pi^{-1}(V)=\pi^{-1}(U\cap V), \end{align*} so the coordinate domain for the transition map is $\varphi(U\cap V)\times\mathbb{R}^n$. Define the base transition map \begin{align*} F: \varphi(U\cap V) &\to \psi(U\cap V) \\ x &\mapsto \psi(\varphi^{-1}(x)). \end{align*} This is smooth because $(U,\varphi)$ and $(V,\psi)$ are charts in the smooth atlas of $M$. Now take $(x,a)\in \varphi(U\cap V)\times\mathbb{R}^n$. The point $x$ represents the base point \begin{align*} p=\varphi^{-1}(x)\in U\cap V, \end{align*} and the vector $a\in\mathbb{R}^n$ represents the tangent vector \begin{align*} v_p=d(\varphi^{-1})_x(a)\in T_pM. \end{align*} Applying the second tangent chart gives \begin{align*} (\widetilde{\psi}\circ \widetilde{\varphi}^{-1})(x,a) &= \widetilde{\psi}(v_p) \\ &= \bigl(\psi(p), d\psi_p(v_p)\bigr) \\ &= \bigl(\psi(\varphi^{-1}(x)), d\psi_{\varphi^{-1}(x)}(d(\varphi^{-1})_x(a))\bigr). \end{align*} The chain rule identifies the second component: \begin{align*} d\psi_{\varphi^{-1}(x)}\circ d(\varphi^{-1})_x=d(\psi\circ\varphi^{-1})_x=dF_x. \end{align*} Therefore \begin{align*} (\widetilde{\psi}\circ \widetilde{\varphi}^{-1})(x,a)=\bigl(F(x),dF_x(a)\bigr). \end{align*} Under the standard coordinate identification of linear maps $\mathbb{R}^n\to\mathbb{R}^n$ with matrices, $dF_x$ is represented by the Jacobian matrix $JF_x$. Thus the transition map is \begin{align*} (x,a)\mapsto \bigl(F(x),JF_xa\bigr). \end{align*} Its first $n$ components are the smooth functions $F_i(x)$. Its last $n$ components are \begin{align*} (x,a)\mapsto \sum_{j=1}^n \partial_{x_j}F_i(x)\,a_j, \end{align*} for $1\le i\le n$. These are smooth because $F$ is smooth, so all partial derivatives $\partial_{x_j}F_i$ are smooth, and multiplication by the coordinate function $a_j$ preserves smoothness. Hence the transition map is smooth. Now define the reverse base transition map \begin{align*} G: \psi(U\cap V) &\to \varphi(U\cap V) \\ y &\mapsto \varphi(\psi^{-1}(y)). \end{align*} This map is smooth because it is the transition map from the chart $(V,\psi)$ to the chart $(U,\varphi)$. For $(y,b)\in \psi(U\cap V)\times\mathbb{R}^n$, let $q=\psi^{-1}(y)$ and $u_q=d(\psi^{-1})_y(b)\in T_qM$. The same chain rule, now written out for the reverse direction, gives \begin{align*} (\widetilde{\varphi}\circ\widetilde{\psi}^{-1})(y,b) &= \widetilde{\varphi}(u_q) \\ &= \bigl(\varphi(q),d\varphi_q(u_q)\bigr) \\ &= \bigl(\varphi(\psi^{-1}(y)),d\varphi_{\psi^{-1}(y)}(d(\psi^{-1})_y(b))\bigr) \\ &= \bigl(G(y),dG_y(b)\bigr). \end{align*} Under the standard bases, $dG_y$ is represented by the Jacobian matrix $JG_y$, so the reverse transition has formula \begin{align*} (y,b)\mapsto \bigl(G(y),JG_yb\bigr). \end{align*} Its last $n$ components are \begin{align*} (y,b)\mapsto \sum_{j=1}^n \partial_{y_j}G_i(y)\,b_j, \end{align*} for $1\le i\le n$, and these functions are smooth because $G$ is smooth. Therefore the inverse transition map is smooth.[/guided]

[step:Define the smooth structure and verify the manifold axioms]Let $\mathcal{A}_{TM}$ be the collection of all maps $\widetilde{\varphi}$ obtained from smooth charts $(U,\varphi)$ on $M$. The preceding step proves that all transition maps between elements of $\mathcal{A}_{TM}$ are smooth. Since the charts on $M$ cover $M$, the domains $\pi^{-1}(U)$ cover $TM$. Define a subset $O\subset TM$ to be open if, for every $\widetilde{\varphi}\in\mathcal{A}_{TM}$, the set $\widetilde{\varphi}(O\cap\pi^{-1}(U))$ is open in $\varphi(U)\times\mathbb{R}^n$. With this topology, each $\widetilde{\varphi}$ is a homeomorphism from $\pi^{-1}(U)$ onto the open subset $\varphi(U)\times\mathbb{R}^n\subset\mathbb{R}^{2n}$. Thus $\mathcal{A}_{TM}$ is a smooth atlas of dimension $2n$. It remains to record that this topology satisfies the Hausdorff and second-countability requirements. Since $M$ is Hausdorff, if $v_p\in T_pM$ and $w_q\in T_qM$ with $p\neq q$, choose disjoint open neighbourhoods $O_p,O_q\subset M$ of $p$ and $q$. For any open subset $O\subset M$ and any tangent chart $\widetilde{\varphi}:\pi^{-1}(U)\to\varphi(U)\times\mathbb{R}^n$, we have \begin{align*} \widetilde{\varphi}\bigl(\pi^{-1}(O)\cap\pi^{-1}(U)\bigr)=\varphi(O\cap U)\times\mathbb{R}^n, \end{align*} which is open in $\varphi(U)\times\mathbb{R}^n$ because $O\cap U$ is open in $U$ and $\varphi$ is a homeomorphism onto $\varphi(U)$. Hence $\pi^{-1}(O)$ is open in the chart-defined topology. Applying this to $O_p$ and $O_q$, the sets $\pi^{-1}(O_p)$ and $\pi^{-1}(O_q)$ are disjoint open neighbourhoods of $v_p$ and $w_q$. If $p=q$, choose a chart $(U,\varphi)$ with $p\in U$; the distinct points $\widetilde{\varphi}(v_p)$ and $\widetilde{\varphi}(w_p)$ in $\mathbb{R}^{2n}$ have disjoint open neighbourhoods, and their inverse images under $\widetilde{\varphi}$ separate $v_p$ and $w_p$. Hence $TM$ is Hausdorff. Since $M$ is second countable, let $\mathcal{C}=\{C_k:k\in\mathbb{N}\}$ be a countable basis for the topology of $M$. Start with the given smooth atlas $\{(U_\alpha,\varphi_\alpha)\}_{\alpha\in A}$ of $M$. For each basis element $C_k$ for which there exists $\alpha\in A$ with $C_k\subset U_\alpha$, choose one such index $\alpha(k)$ and form the restricted smooth chart $(C_k,\varphi_{\alpha(k)}|_{C_k})$. These restricted charts are countable and cover $M$: if $p\in M$, choose $\alpha\in A$ with $p\in U_\alpha$, then choose $C_k\in\mathcal{C}$ with $p\in C_k\subset U_\alpha$. Relabel this countable covering atlas as $\{(U_k,\varphi_k)\}_{k\in\mathbb{N}}$. Let $\mathcal{B}_{2n}$ be the countable basis of $\mathbb{R}^{2n}$ consisting of open balls with rational centres and rational radii. The family \begin{align*} \left\{\widetilde{\varphi}_k^{-1}\bigl(B\cap(\varphi_k(U_k)\times\mathbb{R}^n)\bigr): k\in\mathbb{N},\ B\in\mathcal{B}_{2n}\right\} \end{align*} is countable and forms a basis for the topology on $TM$. Therefore $TM$ is second countable. Thus $TM$ is a smooth $2n$-manifold.[/step]

[guided]After compatibility has been checked, the smooth structure is obtained by declaring the tangent coordinate maps to be charts. Let $\mathcal{A}_{TM}$ denote the collection of all maps $\widetilde{\varphi}$ constructed from smooth charts $(U,\varphi)$ on $M$. Since every point $v_p\in TM$ lies over some $p\in M$, and since the charts on $M$ cover $M$, some $U$ contains $p$; hence $v_p\in\pi^{-1}(U)$. Therefore these tangent chart domains cover $TM$. We define the topology on $TM$ by using these charts: a set $O\subset TM$ is open exactly when its image in every tangent coordinate chart is open. With this definition, each map \begin{align*} \widetilde{\varphi}: \pi^{-1}(U)\to \varphi(U)\times\mathbb{R}^n \end{align*} is a homeomorphism onto an open subset of $\mathbb{R}^{2n}$. The transition maps are smooth by the previous step, so $\mathcal{A}_{TM}$ is a smooth atlas of dimension $2n$. We also need the topological manifold conditions. First, $TM$ is Hausdorff. Let $v_p\in T_pM$ and $w_q\in T_qM$ be distinct points of $TM$. If $p\neq q$, the Hausdorff property of $M$ gives disjoint open neighbourhoods $O_p,O_q\subset M$ with $p\in O_p$ and $q\in O_q$. We first verify that these inverse images are open in the tangent-bundle topology. If $O\subset M$ is open and $\widetilde{\varphi}:\pi^{-1}(U)\to\varphi(U)\times\mathbb{R}^n$ is a tangent chart, then \begin{align*} \widetilde{\varphi}\bigl(\pi^{-1}(O)\cap\pi^{-1}(U)\bigr)=\varphi(O\cap U)\times\mathbb{R}^n. \end{align*} The set $O\cap U$ is open in $U$, and $\varphi$ is a homeomorphism onto $\varphi(U)$, so $\varphi(O\cap U)$ is open in $\varphi(U)$. Hence $\pi^{-1}(O)$ is open by the definition of the tangent-bundle topology. Applying this to $O_p$ and $O_q$, their inverse images $\pi^{-1}(O_p)$ and $\pi^{-1}(O_q)$ are open in the tangent-bundle topology and are disjoint, so they separate $v_p$ and $w_q$. If $p=q$, choose a smooth chart $(U,\varphi)$ with $p\in U$. Since $\widetilde{\varphi}$ is injective, the coordinate points $\widetilde{\varphi}(v_p)$ and $\widetilde{\varphi}(w_p)$ are distinct in $\mathbb{R}^{2n}$. Euclidean space is Hausdorff, so there are disjoint open sets in $\mathbb{R}^{2n}$ separating them. Pulling these open sets back by $\widetilde{\varphi}$ gives disjoint open neighbourhoods of $v_p$ and $w_p$ in $TM$. Second, $TM$ is second countable. We first justify the countable atlas used on $M$. Let $\mathcal{C}=\{C_k:k\in\mathbb{N}\}$ be a countable basis for the topology of $M$, and let $\{(U_\alpha,\varphi_\alpha)\}_{\alpha\in A}$ be the original smooth atlas. Whenever $C_k\subset U_\alpha$ for at least one $\alpha\in A$, choose one such index $\alpha(k)$ and use the restricted chart $(C_k,\varphi_{\alpha(k)}|_{C_k})$. This gives only countably many charts. It covers $M$ because, for each $p\in M$, some chart domain $U_\alpha$ contains $p$, and the basis property gives $C_k$ with $p\in C_k\subset U_\alpha$. Relabel the resulting countable covering atlas as $\{(U_k,\varphi_k)\}_{k\in\mathbb{N}}$. Let $\mathcal{B}_{2n}$ be the countable basis of $\mathbb{R}^{2n}$ consisting of open balls with rational centres and rational radii. Then the sets \begin{align*} \widetilde{\varphi}_k^{-1}\bigl(B\cap(\varphi_k(U_k)\times\mathbb{R}^n)\bigr), \end{align*} where $k\in\mathbb{N}$ and $B\in\mathcal{B}_{2n}$, form a countable family. They form a basis because every [open set](/page/Open%20Set) in $TM$ is tested in local tangent coordinates, and every open subset of $\mathbb{R}^{2n}$ is a union of rational balls. Thus $TM$ is second countable. Together, local Euclidean charts of dimension $2n$, the Hausdorff property, and second countability show that $TM$ is a smooth $2n$-manifold.[/guided]

[step:Show that the tangent bundle projection is smooth] Let $(U,\varphi)$ be a smooth chart on $M$, and use the tangent chart \begin{align*} \widetilde{\varphi}: \pi^{-1}(U)&\to \varphi(U)\times\mathbb{R}^n \end{align*} on $TM$. In these coordinates, the projection $\pi:TM\to M$ is represented by \begin{align*} \varphi\circ \pi\circ \widetilde{\varphi}^{-1}: \varphi(U)\times\mathbb{R}^n &\to \varphi(U) \\ (x,a) &\mapsto x. \end{align*} This map is the restriction of the linear projection $\mathbb{R}^{2n}\to\mathbb{R}^n$ onto the first $n$ coordinates, hence it is smooth. Since smoothness is local in charts, $\pi:TM\to M$ is smooth. This completes the construction of the canonical smooth structure on $TM$ and proves that the projection is smooth. [/step]