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For a smooth map between manifolds $f:M\to N$, the differential $d{f}_x:T_{x}M\to T_{f(x)}N$ is the linear map obtained by differentiating in charts. Intrinsically it sends the tangent vector of a curve $\gamma$ through $x$ to the tangent vector of $f\circ \gamma$. The chain rule makes it independent of charts and gives $d(g\circ f{)}_x=d{g}_{f(x)}\circ d{f}_x$.

Let $f:M\to N$ be a continuous map between oriented $d$-dimensional manifolds, with $d\ge 1$. Suppose $x$ is an isolated point of the fiber over $y=f(x)$. Choose an open neighborhood $U$ with $U\cap f^{-1}(y)=\{x\}$. The map of pairsinduces a map between rank-one degree-$d$ local homology groups. Relative to the local orientation of a manifold, this map is multiplication by an integer, the local degree. Excision makes it independent of the neighborhood. A smooth map between manifolds with invertible derivative at $x$ has local degree $\mathrm{sign}\det D{f}_x$, by the inverse function theorem and the chosen orientations.

For a smooth map between manifolds $\varphi :\mathcal{M}\to \mathcal{N}$, the four constructions go in the directions dictated by composition and the differential of a smooth map.

The pullback of a smooth function $f\in C^{\infty }(\mathcal{N})$ is ${\varphi }^{\ast }f=f\circ \varphi$, a function on $\mathcal{M}$. The pushforward of a curve $\lambda :I\to \mathcal{M}$ is ${\varphi }_{\ast }\lambda =\varphi \circ \lambda$, a curve in $\mathcal{N}$ with the same parameter interval.

For a tangent vector $V\in T_p\mathcal{M}$, its pushforward is ${\varphi }_{\ast }V=d{\varphi }_p(V)\in T_{\varphi (p)}\mathcal{N}$. Intrinsically, treating a tangent vector as a derivation on functions,For a covector $\eta \in T_{\varphi (p)}^{\ast }\mathcal{N}$, the pullback of a covector is the dual linear map:No inverse map is required. A pushed-forward field for a general map is a field along that map; it need not assign a unique vector to an image point with several preimages.

At $p\in \mathcal{Q}$, the differential of a smooth map defines the pushforward of a vector field byThis is a tangent vector at $\psi (p)$, and varies smoothly as a section of the pullback tangent bundle. In local coordinates its components are $X^i(p){\partial }_i{\psi }^a(p)$.

For a general smooth map between manifolds, this is a vector field along a map, rather than an intrinsic vector field on all of $N$. A vector field $Y$ on the image can be defined only if these vectors agree whenever two points have the same image, and the resulting field must also be smooth; a projectable vector field must meet that requirement. For a diffeomorphism, there is no ambiguity and $Y_q=d{\psi }_{{\psi }^{-1}(q)}{X}_{{\psi }^{-1}(q)}$. For example, $\psi (x)=x^2$ on $\mathbb{R}$ and $X={\partial }_x$ give $2x{\partial }_y$, with opposite nonzero values at the two preimages of $y>0$. Thus the source's notation must be read with this qualification.

An embedded smooth $d$-dimensional manifold $X\subset {\mathbb{R}}^N$ is locally carried by a smooth ambient coordinate change to ${\mathbb{R}}^d\times \{0\}$. A map between such manifolds is a smooth map between manifolds if its coordinate representatives are smooth. Its differential $d{f}_x:T_{x}X\to T_{f(x)}Y$ sends the tangent vector represented by a curve $\gamma$ to $(f\circ \gamma {)}^{\prime }(0)$; equivalently it is the derivative in charts.

A regular value $y$ is one for which $d{f}_x$ is surjective at every $x\in f^{-1}(y)$, including vacuous regularity when the fibre is empty. For example zero is not a regular value of $f:\mathbb{R}\to \mathbb{R}$, $f(x)=x^2$, because $d{f}_0=0$.

For an invertible matrix $A$, the derivative of the determinant isAt a determinant-one matrix it is nonzero, since taking $H=A/n$ gives value one. The regular level set theorem therefore makes $S{L}_n(\mathbb{R})={\det }^{-1}(1)$ an embedded manifold of dimension $n^2-1$, also for $n=1$ when it is a single point. Matrix multiplication has polynomial ambient coordinates and restricts to the group, so it is smooth. At the identity,Finally suppose the singular-matrix set were an embedded smooth manifold near zero for $n\ge 2$. Every curve $t\mapsto t{E}_{ij}$ lies in it, since its rank is at most one. Thus every matrix unit belongs to its tangent space at zero. These span ${\mathbb{R}}^{n^2}$, forcing its local dimension to be $n^2$. An embedded manifold of full ambient dimension is locally open, but arbitrarily small scalar multiples of the identity are invertible. This contradiction proves the singular-matrix set is not an embedded smooth manifold. The argument addresses the subset-manifold convention specified at the start.

An abstract smooth surface is a Hausdorff second-countable topological space with an atlas of charts to open subsets of ${\mathbb{R}}^2$ whose transition maps are smooth. It is orientable when it has such an atlas for which every transition map has positive Jacobian determinant. A map $f:S_1\to S_2$ is a smooth map when every coordinate representationis smooth wherever it is defined.

The map $a:C\to C$ is a smooth involution with no fixed point: the equationwould force $x=y=0$, which is impossible on $x^2+y^2=1$. For each $p\in C$, choose a sufficiently small coordinate neighbourhood $U_p$ such thatThen the quotient projection restricts to a homeomorphismTransporting a smooth chart ${\varphi }_p:U_p\to {\mathbb{R}}^2$ across this homeomorphism gives the quotient chartOn an overlap, a lift lies either in another chosen neighbourhood or in its image under $a$. The corresponding transition map is therefore a transition map on $C$, possibly composed with the diffeomorphism $a$, and is smooth. Since this is a free action of the finite group $\{1,a\}$, the quotient is Hausdorff and second countable. This constructs the smooth quotient by a free finite group action, and in these charts $\pi$ is locally the identity. Hence $\pi$ is a local diffeomorphism, in particular smooth.

To test orientability, parametrize the cylinder byThe involution acts in these coordinates aswhose derivative has determinant $-1$. Thus $a$ is an orientation-reversing diffeomorphism of the cylinder.

If the quotient surface $S$ were orientable, its orientation would pull back through the local diffeomorphism $\pi$ to an orientation of $C$. The identity $\pi \circ a=\pi$ would then force $a$ to preserve that pulled-back orientation, contradicting the negative determinant above. Therefore

Define the smooth mapAt $(M,v)$ its derivative in the direction $(A,w)$ isThe restriction to variations $(0,w)$ is the surjective linear map $M:{\mathbb{R}}^n\to {\mathbb{R}}^m$, since $M$ has full row rank. Thus $0$ is a regular value of $\Phi$. The regular level set theorem now shows thatis an embedded submanifold of $X_{m,n}\times {\mathbb{R}}^n$, of dimension $mn+n-m$.

For $x\in S^2\subset {\mathbb{R}}^3$, let $F(x)=x^T$, viewed as a nonzero $1$ by $3$ matrix. This defines a smooth map $F:S^2\to X_{1,3}$. Its pulled-back fiber isso the evident fiberwise identity gives an isomorphism of vector bundles $F^{\ast }{E}_{1,3}\cong T{S}^2$.

If $E_{1,3}$ were a trivial vector bundle, its pullback would be trivial. A trivial rank-two bundle has a nowhere-zero section, whereas the assumed form of the Hairy ball theorem says that the tangent bundle $T{S}^2$ does not. Hence $E_{1,3}$ is nontrivial.

A submersion is a smooth map between manifolds $F:X^n\to Y^m$ for whichis surjective at every $p\in X$.

The local submersion theorem says that around every $p\in X$ there are coordinates $(x^1,\dots ,x^n)$ centered at $p$ and $(y^1,\dots ,y^m)$ centered at $F(p)$ in whichTo prove it, surjectivity lets us choose $m$ source coordinates such that $d{F}^1,\dots ,d{F}^m$ are independent. Complete them by source coordinates $x^{m+1},\dots ,x^n$ and defineThe derivative of $G$ is invertible at $p$, so the inverse function theorem makes $G$ a local coordinate system; in these coordinates $F$ is the displayed projection.

For $q\in Y$, the fiber is locally given byThese are slice coordinates, so $F^{-1}(q)$ is an embedded submanifold of dimension $n-m$.

A vector field $X$ is projectable through a smooth map between manifolds $f:M\to N$ if there is a smooth vector field $Y$ on $N$ with $d{f}_p{X}_p=Y_{f(p)}$ for every $p$. Agreement on fibres is necessary. For a surjective submersion it is also sufficient: smooth local sections of the submersion express $Y$ locally as $df(X)$ and prove its smoothness. An arbitrary non-surjective map may instead give a field only along its image, with extension to $N$ requiring additional choices.

A pullback of a smooth function composes a function on the target with a smooth map between manifolds, producing a function on the source. It needs no inverse map. Acting on the pulled-back function defines the differential of a smooth map intrinsically as a map of derivations.

For a smooth map between manifolds $f:M\to N$, the pullback vector bundle $f^{\ast }TN$ has fibre $T_{f(p)}N$ at $p\in M$. A smooth section is a vector field along a map. It allows one to differentiate maps and curves without requiring them to be invertible.

A pushforward of a curve composes a parametrized curve in the source with a smooth map between manifolds. Its tangent is the image of the original tangent under the differential of a smooth map. A constant image curve has zero tangent, so a general smooth map need not preserve regularity of curves.

For a vector field $X$ on $M$, its pushforward by a smooth map between manifolds $f:M\to N$ is the vector field along a map defined by $(f_{\ast }X{)}_p=d{f}_p(X_p)$. Acting on a function $h$ on $N$, it satisfies $(f_{\ast }X{)}_p(h)=X_p(h\circ f)$. For a diffeomorphism this defines a vector field on $N$ by evaluation at $p=f^{-1}(q)$. For a general map there need not be a single vector at each image point: $f(x)=x^2$, $X={\partial }_x$ gives $2x{\partial }_y$, with opposite values over $y>0$.

A vector field along a smooth map between manifolds $f:M\to N$ assigns $Y_p\in T_{f(p)}N$ smoothly to each $p$. It is a section of the pullback tangent bundle, not necessarily a vector field on $N$. Along a curve $\gamma$, the velocity $\dot{\gamma }$ is such a section.