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For a positive-dimensional smooth connected manifold, three equivalent formulations of orientability of a smooth manifold are an oriented atlas, a continuously varying orientation of each tangent space, and a nowhere-vanishing smooth top-degree differential form. An orientation of a tangent space means a choice of one class of ordered bases, where two bases agree when their change-of-basis determinant is positive. The second formulation chooses these classes locally continuously; it does not require a global frame.

An oriented atlas has positive Jacobian determinant on every overlap. Its coordinate bases then give a consistent orientation of all tangent spaces. Conversely, a continuous choice of tangent-space orientations makes the sign of a coordinate basis locally constant. Shrink to connected charts and reverse one coordinate when necessary; the resulting positively positively oriented coordinate charts have positive transition determinants. An orientation in atlas language is the maximal collection of charts compatible with that positive sign convention.

We use the partition of unity theorem: every open cover of a Hausdorff second-countable smooth manifold has a smooth locally finite subordinate partition. From the oriented atlas, choose such a partition of unity ${\rho }_i$. Extend each ${\rho }_{i}d{x}_i^1\wedge \cdots \wedge d{x}_i^n$ by zero outside its chart and sum the locally finite family. At every point, the nonzero summands are positive multiples of one another and at least one coefficient is positive. Their sum is a smooth nowhere-zero form $\omega$. Conversely, a nowhere-zero form declares a basis positive exactly when $\omega (v_1,\dots ,v_n)>0$, giving a continuously varying orientation and hence an oriented atlas. Two top forms define the same orientation precisely when ${\omega }^{\prime }=h\omega$ for a smooth positive function $h$. Since a nonvanishing coefficient cannot change sign on a connected manifold, these definitions give exactly two opposite orientations.

For the unit sphere in ${\mathbb{R}}^{n+1}$, defineThe radial vector $x$ is a unit normal, so appending any tangent basis gives an ambient basis and the determinant never vanishes. The form is smooth and provides the outward-normal-first orientation. Therefore $\boxed{S^n\text{ is orientable for every }n}$. For $n=0$, this is the nonzero zero-form taking opposite signs on the two sphere points; zero-dimensional manifolds admit orientations by assigning generator signs at their points.

For $n\ge 1$, the quotient map $\pi :S^n\to {\mathbb{RP}}^n$ is a twofold smooth covering with deck transformation $A(x)=-x$. Its action on the sphere form isIf $n$ is odd, the form is invariant and descends using the local covering inverses to a nowhere-vanishing form on the quotient. If $n$ is even, suppose the quotient had a nonvanishing top form $\eta$. Its pullback would be $h\omega$ for a smooth nonzero function $h$ on the connected sphere and would be invariant under $A$. Invariance would require $h(-x)=-h(x)$, impossible because $h$ has constant sign. This proves the orientability of real projective space criterionThe separate endpoint ${\mathbb{RP}}^0$ is one point and is orientable, so the full nonnegative-dimensional answer is $n=0$ or $n$ odd.

A $d$-dimensional embedded submanifold $F\subseteq M^m$ has the subspace topology and, near each of its points, a manifold chart identifying it with a coordinate slice ${\mathbb{R}}^d\times \{0\}$. These slice charts for an embedded submanifold give its smooth structure, and its inclusion is a smooth embedding.

A regular value $q\in N^n$ of a smooth map $f:M^m\to N^n$ is one for which $d{f}_p:T_{p}M\to T_{q}N$ is surjective at every $p\in f^{-1}(q)$. If the preimage is nonempty this forces $m\ge n$.

The inverse function theorem used here says that if $G:U\subseteq {\mathbb{R}}^m\to {\mathbb{R}}^m$ is smooth and $D{G}_p$ is an invertible matrix, then its restriction to a suitable neighbourhood of $p$ is a diffeomorphism onto an open neighbourhood of $G(p)$, with smooth inverse.

To prove the regular level set theorem, fix $p\in f^{-1}(q)$ and choose coordinates centred at $p,q$. The coordinate representative $\tilde{f}$ has a rank-$n$ Jacobian matrix. After permuting domain coordinates, its first $n$ columns form an invertible matrix. DefineIts Jacobian matrix is block triangular with invertible diagonal blocks, so the inverse function theorem makes $G$ a local coordinate change. In these coordinates, $f^{-1}(q)$ is precisely the slice with first $n$ coordinates zero. Thus it is an embedded submanifold of dimension $m-n$, withAn empty preimage is harmless; the local assertion is then vacuous.

Three equivalent formulations of orientability of a smooth manifold are: an oriented atlas whose transition Jacobian determinants are positive; a smoothly varying orientation of a vector space on each tangent space; and a nowhere-vanishing smooth top-degree differential form. An oriented atlas gives compatible oriented coordinate frames. A partition of unity glues their positive local top forms, and a nonzero top form declares a frame positive precisely when it evaluates positively on that frame. These constructions prove the equivalence.

For the orientability of a regular fibre, write $F=f^{-1}(q)$. The derivative gives a short exact sequence of vector bundlesThis is the reason the normal bundle of a regular fibre is trivial. Choose any fixed positive basis $b_1,\dots ,b_n$ of $T_{q}N$ and an ambient orientation on $M$. Declare a frame $v_1,\dots ,v_{m-n}$ of $T_{p}F$ positive whenis positive in $T_{p}M$, where $d{f}_p({\tilde{b}}_i)=b_i$. Changing a lift adds a tangent vector and leaves the determinant sign unchanged. Local smooth lifts exist because $df$ is surjective, so this rule varies smoothly and agrees on overlaps. It gives a global orientation on $F$. Only an orientation of the single vector space $T_{q}N$ was chosen; $N$ itself need not be orientable.

An orientation of a smooth $n$-dimensional manifold can be specified by a nowhere-vanishing smooth $n$-form $\omega$: an ordered tangent basis is positive precisely when $\omega$ evaluates positively on it. Multiplication by a positive smooth function preserves the orientation. This is the differential-form formulation of the equivalent formulations of orientability of a smooth manifold.

Use the standard convention that the compact connected manifold is nonempty and has no boundary. Choose a positive volume form $\omega$ for its orientation. It is a closed differential form, because there are no nonzero $(n+1)$-forms. For $n>0$, compactness makes its integral finite, and positivity on a coordinate neighborhood together with nonnegativity elsewhere givesIf $\omega =d\eta$, the Generalized Stokes theorem would give ${\int }_M\omega ={\int }_{\partial M}\eta =0$, a contradiction. ThereforeFor $n=0$, a nonempty connected manifold is a point, and $H_{\mathrm{dR}}^0(M)=\mathbb{R}$ directly. The boundary hypothesis matters: a closed interval is an oriented compact connected manifold with boundary with $H_{\mathrm{dR}}^1=0$. Thus the result is about ordinary boundaryless manifolds, not an unrestricted version including boundary.

The equivalent formulations of orientability of a smooth manifold are:

A positive chart orients its coordinate frame. Conversely, smoothly oriented frames determine local positive coordinate volume forms. A partition of unity subordinate to an oriented atlas glues these positive forms: at each point they are positive multiples of one another, so their weighted sum cannot vanish. Finally, a nowhere-zero top form $\omega$ declares $(v_1,\dots ,v_n)$ positive exactly when $\omega (v_1,\dots ,v_n)>0$. These constructions are inverse at the level of orientations.