[proofplan] The argument has two stages. First, the [Dimensions of Symmetric and Exterior Powers](/theorems/3303) gives $\dim \Lambda^n(V^*) = 1$, so any nonzero $\omega_0 \in \Lambda^n(V^*)$ induces a linear isomorphism $\mathbb{R} \to \Lambda^n(V^*)$, $t \mapsto t\omega_0$, which is a homeomorphism in the canonical topology. Transporting the [decomposition of $\mathbb{R} \setminus \{0\}$ into its two open intervals](/theorems/295) along this homeomorphism yields exactly two connected components $C^+ = \mathbb{R}_{>0}\,\omega_0$ and $C^- = \mathbb{R}_{<0}\,\omega_0$. Second, given any ordered basis $(e_1, \dots, e_n)$ of $V$ with [dual basis](/theorems/414) $(e^1, \dots, e^n)$, the form $e^1 \wedge \cdots \wedge e^n$ is a nonzero element of $\Lambda^n(V^*)$; under a [change of basis](/theorems/3275) with matrix $A$, the new top form equals $(\det A)^{-1}$ times the old, so two ordered bases land in the same component iff their change-of-basis determinant is positive iff they define the same orientation. [/proofplan] [step:Establish that $\Lambda^n(V^*)$ is one-dimensional and inherits the standard topology of $\mathbb{R}$] By the [Dimensions of Symmetric and Exterior Powers](/theorems/3303) applied to $V^*$, which has $\dim V^* = n$, we obtain \begin{align*} \dim_{\mathbb{R}} \Lambda^n(V^*) = \binom{n}{n} = 1. \end{align*} Fix any nonzero element $\omega_0 \in \Lambda^n(V^*)$ — such an element exists because $\Lambda^n(V^*)$ has positive dimension, and explicitly, choosing any ordered basis $(e_1,\dots,e_n)$ of $V$ with [dual basis](/theorems/414) $(e^1,\dots,e^n) \in V^*$, by the [Basis of Exterior Powers of the Dual Space](/theorems/3561) the single wedge $e^1 \wedge \cdots \wedge e^n$ is a basis of $\Lambda^n(V^*)$, hence nonzero. Define the map \begin{align*} \Psi: \mathbb{R} &\to \Lambda^n(V^*) \\ t &\mapsto t\,\omega_0. \end{align*} Since $\omega_0$ spans $\Lambda^n(V^*)$, $\Psi$ is a linear bijection. As a linear bijection between finite-dimensional real vector spaces, $\Psi$ is a homeomorphism with respect to the unique Hausdorff topological-vector-space topology on each side (the standard topology on $\mathbb{R}$, and the topology on $\Lambda^n(V^*)$ obtained by transport of structure from any linear isomorphism with $\mathbb{R}^1$). [guided] Why does it suffice to fix one nonzero $\omega_0$? Because $\Lambda^n(V^*)$ is one-dimensional, every nonzero element is a basis, and every other element is a unique real multiple of it. So the map $t \mapsto t\omega_0$ identifies $\Lambda^n(V^*)$ with $\mathbb{R}$ in the most rigid way possible — linearly, bijectively, and continuously in both directions. The [Dimensions of Symmetric and Exterior Powers](/theorems/3303) gives the dimension formula $\dim \Lambda^k(W) = \binom{\dim W}{k}$ for a [vector space](/page/Vector%20Space) $W$ and $0 \le k \le \dim W$. With $W = V^*$ (so $\dim W = n$) and $k = n$, this is $\binom{n}{n} = 1$, as claimed. To exhibit a concrete nonzero element, we pick an ordered basis $(e_1,\dots,e_n)$ of $V$, take its [dual basis](/theorems/414) $(e^1,\dots,e^n)$ of $V^*$ (defined by $e^i(e_j) = \delta_{ij}$), and form the wedge $e^1 \wedge \cdots \wedge e^n$. The [Basis of Exterior Powers of the Dual Space](/theorems/3561) tells us that as the multi-index $1 \le i_1 < \cdots < i_n \le n$ runs over strictly increasing $n$-tuples, the wedges $e^{i_1} \wedge \cdots \wedge e^{i_n}$ form a basis of $\Lambda^n(V^*)$. The only such multi-index is $(1,2,\dots,n)$, confirming both that $\Lambda^n(V^*)$ is one-dimensional and that $e^1 \wedge \cdots \wedge e^n \ne 0$. Finally, on any finite-dimensional real [vector space](/page/Vector%20Space) there is a unique topology making it a Hausdorff [topological vector space](/page/Topological%20Vector%20Space) (this topology is induced by any norm, and all norms are equivalent). The linear bijection $\Psi$ pulls the standard topology of $\mathbb{R}$ back to this canonical topology on $\Lambda^n(V^*)$, so $\Psi$ is automatically a homeomorphism: both $\Psi$ and $\Psi^{-1}$ are linear maps between one-dimensional real vector spaces, and every [linear map](/page/Linear%20Map) between finite-dimensional normed spaces is continuous. [/guided] [/step] [step:Conclude that $\Lambda^n(V^*) \setminus \{0\}$ has exactly two connected components] The restriction \begin{align*} \Psi\big|_{\mathbb{R} \setminus \{0\}}: \mathbb{R} \setminus \{0\} \to \Lambda^n(V^*) \setminus \{0\} \end{align*} is a homeomorphism, since $\Psi$ is a homeomorphism and maps $0$ to $0$ (so it restricts to a homeomorphism of complements). By the [classification of connected subsets of $\mathbb{R}$](/theorems/295), the connected subsets of $\mathbb{R}$ are exactly the intervals. The set $\mathbb{R} \setminus \{0\} = (-\infty, 0) \cup (0, \infty)$ is a disjoint union of two open intervals, each of which is connected. Neither $(-\infty,0)$ nor $(0,\infty)$ can be enlarged within $\mathbb{R} \setminus \{0\}$ to a larger connected set, because any subset of $\mathbb{R} \setminus \{0\}$ containing points on both sides of $0$ would fail to be an interval. Hence $\mathbb{R} \setminus \{0\}$ has exactly two connected components, $(-\infty,0)$ and $(0,\infty)$. By the [preservation of connected components under homeomorphisms](/theorems/302), the homeomorphism $\Psi|_{\mathbb{R}\setminus\{0\}}$ sends connected components to connected components bijectively. Therefore $\Lambda^n(V^*) \setminus \{0\}$ has exactly two connected components, \begin{align*} C^+ &:= \Psi\bigl((0,\infty)\bigr) = \{t\,\omega_0 : t > 0\}, \\ C^- &:= \Psi\bigl((-\infty,0)\bigr) = \{t\,\omega_0 : t < 0\}. \end{align*} [guided] The point here is purely topological. Once we have identified $\Lambda^n(V^*)$ with $\mathbb{R}$ via the homeomorphism $\Psi$, the question "how many connected components does $\Lambda^n(V^*) \setminus \{0\}$ have?" becomes "how many connected components does $\mathbb{R} \setminus \{0\}$ have?", because homeomorphisms preserve the connected-component decomposition. Why are there exactly two components in $\mathbb{R} \setminus \{0\}$? The [classification of connected subsets of $\mathbb{R}$](/theorems/295) states that a subset $S \subseteq \mathbb{R}$ is connected iff $S$ is an interval (possibly empty, a singleton, or unbounded). The sets $(-\infty,0)$ and $(0,\infty)$ are intervals, hence connected. To see they are maximal connected subsets of $\mathbb{R}\setminus\{0\}$: any connected subset $S$ of $\mathbb{R}\setminus\{0\}$ is an interval, and an interval that contains a negative number $a$ and a positive number $b$ must contain $[a,b]$, hence $0$ — contradicting $S \subseteq \mathbb{R}\setminus\{0\}$. So no connected subset of $\mathbb{R}\setminus\{0\}$ straddles $0$, which means $(-\infty,0)$ and $(0,\infty)$ are precisely the two connected components. Now invoke [preservation of connected components under homeomorphisms](/theorems/302): for any homeomorphism $h: X \to Y$, the image of a connected component of $X$ is a connected component of $Y$, and this correspondence is bijective. Applying this to $h = \Psi|_{\mathbb{R}\setminus\{0\}}$, the two components of $\mathbb{R}\setminus\{0\}$ correspond bijectively to two components $C^+$, $C^-$ of $\Lambda^n(V^*)\setminus\{0\}$. These are the open half-lines $\mathbb{R}_{>0}\,\omega_0$ and $\mathbb{R}_{<0}\,\omega_0$ through the chosen reference form $\omega_0$. [/guided] [/step] [step:Encode an ordered basis as a top form via its dual basis] For an ordered basis $\mathcal{e} = (e_1, \dots, e_n)$ of $V$, let $(e^1, \dots, e^n) \in V^*$ denote its [dual basis](/theorems/414), characterised by $e^i(e_j) = \delta_{ij}$. Define the top form associated with $\mathcal{e}$: \begin{align*} \omega_{\mathcal{e}} &:= e^1 \wedge e^2 \wedge \cdots \wedge e^n \in \Lambda^n(V^*). \end{align*} By the [Basis of Exterior Powers of the Dual Space](/theorems/3561) (the special case where the strictly increasing multi-index ranges only over $(1,2,\dots,n)$), $\omega_{\mathcal{e}}$ is a basis of the one-dimensional space $\Lambda^n(V^*)$. In particular $\omega_{\mathcal{e}} \ne 0$, so $\omega_{\mathcal{e}} \in \Lambda^n(V^*) \setminus \{0\}$. Now let $\mathcal{f} = (f_1, \dots, f_n)$ be a second ordered basis of $V$, and let $A = (A_{ij}) \in \mathbb{R}^{n\times n}$ be the change-of-basis matrix from $\mathcal{e}$ to $\mathcal{f}$, defined by \begin{align*} f_j = \sum_{i=1}^n A_{ij}\,e_i \qquad (j = 1, \dots, n). \end{align*} By the [Change of Basis Transformation](/theorems/3275) in the [dual space](/page/Dual%20Space) (see in particular [Change Of Basis In Dual Space](/theorems/416)), the dual bases transform by the inverse matrix in the present indexing convention: writing $B = A^{-1}$, one has \begin{align*} f^i = \sum_{k=1}^n B_{ik}\,e^k \qquad (i = 1, \dots, n). \end{align*} Indeed, applying both sides to $f_j = \sum_{l=1}^n A_{lj}\,e_l$ gives \begin{align*} \delta_{ij} = f^i(f_j) = \sum_{k=1}^n B_{ik}A_{kj} = (BA)_{ij}, \end{align*} so $BA = I_n$ and $B = A^{-1}$. By the multilinearity and alternation of the wedge product, \begin{align*} \omega_{\mathcal{f}} = f^1 \wedge \cdots \wedge f^n = \det(B)\,(e^1 \wedge \cdots \wedge e^n) = \det(B)\,\omega_{\mathcal{e}}. \end{align*} Since $B = A^{-1}$, we have $\det B = \det(A^{-1}) = (\det A)^{-1}$. Therefore \begin{align*} \boxed{\;\omega_{\mathcal{f}} = (\det A)^{-1}\,\omega_{\mathcal{e}}.\;} \end{align*} [guided] The map $\mathcal{e} \mapsto \omega_{\mathcal{e}}$ is the bridge between bases of $V$ and top forms: it converts the algebraic notion of an "ordered basis up to $GL$" into the analytic notion of a "nonzero top form up to scalar". We need two facts about this bridge: (a) Every $\omega_{\mathcal{e}}$ is nonzero. This is exactly what the [Basis of Exterior Powers of the Dual Space](/theorems/3561) guarantees: the wedges $e^{i_1} \wedge \cdots \wedge e^{i_k}$ over strictly increasing multi-indices form a basis of $\Lambda^k(V^*)$, and for $k = n$ the only such multi-index is $(1,2,\dots,n)$, so $\omega_{\mathcal{e}}$ alone is a basis of the one-dimensional space $\Lambda^n(V^*)$. (b) Under a change of basis $f_j = \sum_i A_{ij} e_i$, the top form transforms by $\omega_{\mathcal{f}} = (\det A)^{-1}\omega_{\mathcal{e}}$. The crucial computation is the wedge expansion: \begin{align*} \omega_{\mathcal{f}} = f^1 \wedge \cdots \wedge f^n = \sum_{k_1,\dots,k_n=1}^n B_{1\,k_1}\cdots B_{n\,k_n}\,e^{k_1} \wedge \cdots \wedge e^{k_n}. \end{align*} Because the wedge product is alternating, $e^{k_1}\wedge\cdots\wedge e^{k_n} = 0$ unless $(k_1,\dots,k_n)$ is a permutation $\sigma$ of $(1,\dots,n)$, in which case it equals $\operatorname{sgn}(\sigma)\,(e^1\wedge\cdots\wedge e^n)$. Collecting, \begin{align*} \omega_{\mathcal{f}} = \Bigl(\sum_{\sigma \in S_n} \operatorname{sgn}(\sigma)\,B_{1\,\sigma(1)}\cdots B_{n\,\sigma(n)}\Bigr)\,\omega_{\mathcal{e}} = \det(B)\,\omega_{\mathcal{e}}. \end{align*} Since the [dual basis](/theorems/414) transforms by the inverse matrix in the present convention ([Change Of Basis In Dual Space](/theorems/416)), $B = A^{-1}$. The indexing can also be checked directly: applying $f^i = \sum_{k=1}^n B_{ik}\,e^k$ to $f_j = \sum_{l=1}^n A_{lj}\,e_l$ gives \begin{align*} \delta_{ij} = f^i(f_j) = \sum_{k=1}^n B_{ik}A_{kj} = (BA)_{ij}, \end{align*} so $BA = I_n$ and $B = A^{-1}$. Hence $\det B = \det(A^{-1}) = (\det A)^{-1}$. The factor $(\det A)^{-1}$ — rather than $\det A$ — is a feature of the *dual* transformation, but for the purposes of detecting the *sign* it is immaterial: $\det A$ and $(\det A)^{-1}$ have the same sign. [/guided] [/step] [step:Identify the two components with the two orientations of $V$] Recall that an *orientation* of $V$ is, by definition, an equivalence class of ordered bases under the relation \begin{align*} \mathcal{e} \sim \mathcal{f} \iff \det A > 0, \end{align*} where $A$ is the change-of-basis matrix from $\mathcal{e}$ to $\mathcal{f}$. This relation has exactly two equivalence classes whenever $n \ge 1$: indeed, fix any basis $\mathcal{e}$; then the class of $\mathcal{e}$ contains exactly those bases related to $\mathcal{e}$ by a positive-determinant change, and the class of $(-e_1, e_2, \dots, e_n)$ contains all the others (since $\det A < 0$ for any $A$ with negative determinant, and composing with the matrix flipping the first vector reverses the sign of the determinant). Define the map \begin{align*} \Theta: \{\text{orientations of }V\} &\to \{\text{connected components of }\Lambda^n(V^*)\setminus\{0\}\} \\ [\mathcal{e}] &\mapsto \text{the component containing }\omega_{\mathcal{e}}. \end{align*} **Well-defined.** Suppose $\mathcal{e} \sim \mathcal{f}$, i.e.\ the change-of-basis matrix $A$ satisfies $\det A > 0$. From the previous step, $\omega_{\mathcal{f}} = (\det A)^{-1}\omega_{\mathcal{e}}$ with $(\det A)^{-1} > 0$. Writing $\omega_{\mathcal{e}} = t_{\mathcal{e}}\,\omega_0$ for some $t_{\mathcal{e}} \in \mathbb{R} \setminus \{0\}$, we then have $\omega_{\mathcal{f}} = (\det A)^{-1}\,t_{\mathcal{e}}\,\omega_0$, which lies on the same side of $0$ as $t_{\mathcal{e}}\,\omega_0$ since the scaling factor is positive. Hence $\omega_{\mathcal{e}}$ and $\omega_{\mathcal{f}}$ belong to the same connected component (the same one of $C^+$, $C^-$). **Injective.** Conversely, suppose $\omega_{\mathcal{e}}$ and $\omega_{\mathcal{f}}$ lie in the same component. Writing $\omega_{\mathcal{e}} = t_{\mathcal{e}}\omega_0$ and $\omega_{\mathcal{f}} = t_{\mathcal{f}}\omega_0$, this means $t_{\mathcal{e}}$ and $t_{\mathcal{f}}$ have the same sign, so $t_{\mathcal{f}}/t_{\mathcal{e}} > 0$. From $\omega_{\mathcal{f}} = (\det A)^{-1}\omega_{\mathcal{e}}$, we read off $(\det A)^{-1} = t_{\mathcal{f}}/t_{\mathcal{e}} > 0$, hence $\det A > 0$, hence $\mathcal{e} \sim \mathcal{f}$. **Surjective.** Each of $C^+$ and $C^-$ contains at least one $\omega_{\mathcal{e}}$. For $C^+$: take any ordered basis $\mathcal{e}$; if $\omega_{\mathcal{e}} \in C^+$ we are done, otherwise $\omega_{\mathcal{e}} \in C^-$ and we replace $\mathcal{e}$ by $\mathcal{e}' = (-e_1, e_2, \dots, e_n)$. The change-of-basis matrix from $\mathcal{e}$ to $\mathcal{e}'$ has determinant $-1$, so by the formula of the previous step $\omega_{\mathcal{e}'} = (-1)^{-1}\omega_{\mathcal{e}} = -\omega_{\mathcal{e}} \in C^+$. The argument for $C^-$ is symmetric. Therefore $\Theta$ is a bijection between the set of orientations of $V$ and the set of connected components of $\Lambda^n(V^*) \setminus \{0\}$. This completes the proof. [guided] We now assemble all the pieces, and we repeat the set-theoretic checks because this is where the statement identifies topology with orientation. An orientation of $V$ is the following equivalence class of ordered bases: for ordered bases $\mathcal{e} = (e_1,\dots,e_n)$ and $\mathcal{f} = (f_1,\dots,f_n)$, define \begin{align*} \mathcal{e} \sim \mathcal{f} \iff \det A > 0, \end{align*} where $A = (A_{ij}) \in \mathbb{R}^{n\times n}$ is the change-of-basis matrix from $\mathcal{e}$ to $\mathcal{f}$, defined by \begin{align*} f_j = \sum_{i=1}^n A_{ij}\,e_i \qquad (j = 1, \dots, n). \end{align*} This relation has exactly two equivalence classes when $n \ge 1$. To see this, fix an ordered basis $\mathcal{e} = (e_1,\dots,e_n)$. If $\mathcal{f}$ has change-of-basis matrix $A$ from $\mathcal{e}$ and $\det A > 0$, then $\mathcal{f}$ lies in the class of $\mathcal{e}$. If $\det A < 0$, let $R \in \mathbb{R}^{n\times n}$ be the diagonal matrix with entries $R_{11} = -1$ and $R_{ii} = 1$ for $2 \le i \le n$, and let $\mathcal{e}' = (-e_1,e_2,\dots,e_n)$. The change-of-basis matrix from $\mathcal{e}$ to $\mathcal{e}'$ is $R$, so $\det R = -1$. The change-of-basis matrix from $\mathcal{e}'$ to $\mathcal{f}$ is $R^{-1}A = RA$, and \begin{align*} \det(RA) = \det R\,\det A = -\det A > 0. \end{align*} Thus every ordered basis lies in the class of either $\mathcal{e}$ or $\mathcal{e}'$, and these two classes are distinct because the determinant from $\mathcal{e}$ to $\mathcal{e}'$ is $-1 < 0$. Define the map \begin{align*} \Theta: \{\text{orientations of }V\} &\to \{\text{connected components of }\Lambda^n(V^*)\setminus\{0\}\} \\ [\mathcal{e}] &\mapsto \text{the component containing }\omega_{\mathcal{e}}. \end{align*} We must check that this map is well-defined, injective, and surjective. First, suppose $\mathcal{e} \sim \mathcal{f}$. Let $A \in \mathbb{R}^{n\times n}$ be the change-of-basis matrix from $\mathcal{e}$ to $\mathcal{f}$. Then $\det A > 0$. From the preceding step, the associated top forms satisfy \begin{align*} \omega_{\mathcal{f}} = (\det A)^{-1}\omega_{\mathcal{e}}. \end{align*} Because $(\det A)^{-1} > 0$, multiplication by this scalar preserves the sign of the coordinate in the fixed decomposition through $\omega_0$. More explicitly, since $\omega_{\mathcal{e}} \ne 0$ and $\omega_0$ spans $\Lambda^n(V^*)$, there is a unique $t_{\mathcal{e}} \in \mathbb{R}\setminus\{0\}$ such that \begin{align*} \omega_{\mathcal{e}} = t_{\mathcal{e}}\,\omega_0. \end{align*} Then \begin{align*} \omega_{\mathcal{f}} = (\det A)^{-1}t_{\mathcal{e}}\,\omega_0, \end{align*} and $(\det A)^{-1}t_{\mathcal{e}}$ has the same sign as $t_{\mathcal{e}}$. Therefore $\omega_{\mathcal{e}}$ and $\omega_{\mathcal{f}}$ lie in the same one of the two components $C^+$ and $C^-$. Hence $\Theta$ does not depend on the chosen representative of the orientation class. Next, suppose $\Theta([\mathcal{e}]) = \Theta([\mathcal{f}])$. This means that $\omega_{\mathcal{e}}$ and $\omega_{\mathcal{f}}$ lie in the same connected component of $\Lambda^n(V^*)\setminus\{0\}$. Since $\omega_0$ spans $\Lambda^n(V^*)$, there are unique scalars $t_{\mathcal{e}}, t_{\mathcal{f}} \in \mathbb{R}\setminus\{0\}$ such that \begin{align*} \omega_{\mathcal{e}} &= t_{\mathcal{e}}\omega_0, \\ \omega_{\mathcal{f}} &= t_{\mathcal{f}}\omega_0. \end{align*} Being in the same component means $t_{\mathcal{e}}$ and $t_{\mathcal{f}}$ have the same sign, so \begin{align*} \frac{t_{\mathcal{f}}}{t_{\mathcal{e}}} > 0. \end{align*} If $A$ is the change-of-basis matrix from $\mathcal{e}$ to $\mathcal{f}$, the formula from the previous step gives \begin{align*} \omega_{\mathcal{f}} = (\det A)^{-1}\omega_{\mathcal{e}}. \end{align*} Substituting the two scalar expressions relative to $\omega_0$ gives \begin{align*} t_{\mathcal{f}}\omega_0 = (\det A)^{-1}t_{\mathcal{e}}\omega_0. \end{align*} Since $\omega_0 \ne 0$, we may compare coefficients in the one-dimensional [vector space](/page/Vector%20Space) $\Lambda^n(V^*)$ and obtain \begin{align*} (\det A)^{-1} = \frac{t_{\mathcal{f}}}{t_{\mathcal{e}}} > 0. \end{align*} Therefore $\det A > 0$, so $\mathcal{e} \sim \mathcal{f}$. Thus $\Theta$ is injective. Finally, we prove surjectivity. Let $C^+$ and $C^-$ be the two connected components determined by the fixed reference form $\omega_0$: \begin{align*} C^+ &= \{t\omega_0 : t > 0\}, \\ C^- &= \{t\omega_0 : t < 0\}. \end{align*} Choose any ordered basis $\mathcal{e} = (e_1,\dots,e_n)$ of $V$. Its associated top form $\omega_{\mathcal{e}}$ is nonzero, so it belongs to exactly one of $C^+$ and $C^-$. If $\omega_{\mathcal{e}} \in C^+$, then $C^+$ is hit by $\Theta$. If instead $\omega_{\mathcal{e}} \in C^-$, define $\mathcal{e}' = (-e_1,e_2,\dots,e_n)$. The change-of-basis matrix from $\mathcal{e}$ to $\mathcal{e}'$ is the diagonal matrix $R$ with determinant $-1$, hence the top-form transformation formula gives \begin{align*} \omega_{\mathcal{e}'} = (\det R)^{-1}\omega_{\mathcal{e}} = (-1)^{-1}\omega_{\mathcal{e}} = -\omega_{\mathcal{e}}. \end{align*} Thus $\omega_{\mathcal{e}'} \in C^+$. The same argument also hits $C^-$: if $\omega_{\mathcal{e}} \in C^-$ we are done, while if $\omega_{\mathcal{e}} \in C^+$ then the same sign-flipped basis $\mathcal{e}'$ satisfies $\omega_{\mathcal{e}'} = -\omega_{\mathcal{e}} \in C^-$. Therefore every connected component of $\Lambda^n(V^*)\setminus\{0\}$ is in the image of $\Theta$. We have shown that $\Theta$ is well-defined, injective, and surjective. Consequently the two connected components of $\Lambda^n(V^*)\setminus\{0\}$ correspond exactly to the two orientations of $V$, with an orientation represented by $\mathcal{e}$ sent to the component containing the top form $\omega_{\mathcal{e}}$. This proves the theorem. [/guided] [/step]