Smooth Functor of Points I: Tangent Spaces via Dual Numbers

The Lawvere theory of smooth functionsSmooth manifolds give $C^\infty$-ringsThe smooth functor of pointsThe dual numbers as a $C^\infty$-ringTangent vectors as dual-number pointsThe key verification: dual-number points are point derivationsWhy the formula has this shapeThe $C^\infty$-operation on the dual numbersLemma: point derivations kill constantsLemma: Hadamard lemmaLemma: point derivations satisfy the chain ruleCompletion of the proofThe total tangent bundle as the set of dual-number points

In algebraic geometry, one does not study a space only through its ordinary points. One studies its functor of points.

For an affine scheme

$$X=\mathrm{Spec}A,$$

one defines

$$X(R)={\mathrm{Hom}}_{\mathbf{CAlg}}(A,R).$$

The ordinary points are only the values at fields. Infinitesimal rings such as

$$k[\epsilon ]/({\epsilon }^2)$$

also test the space. In particular, tangent vectors can be described as maps from the dual numbers.

The purpose of this article is to explain the smooth analogue of this picture.

For a smooth manifold $M$$M$, the relevant algebra is not a polynomial ring or a commutative algebra of regular functions. It is the $C^{\mathrm{\infty }}$$C^\infty$-ring

$$C^{\mathrm{\infty }}(M).$$

The corresponding functor of points is

$$M(A)={\mathrm{Hom}}_{C^{\mathrm{\infty }}\mathrm{Ring}}(C^{\mathrm{\infty }}(M),A),$$

where $A$$A$ ranges over $C^{\mathrm{\infty }}$$C^\infty$-rings.

The main point is:

$$\boxed{{\displaystyle TM\cong M(\mathbb{R}[\epsilon ]/({\epsilon }^2)).}}$$

Thus tangent vectors are not first introduced as equivalence classes of curves. They are infinitesimal points over the dual numbers.

The Lawvere theory of smooth functions

Let

$${\mathcal{L}}_{C^{\mathrm{\infty }}}$$

be the Lawvere theory of smooth functions.

Its objects are natural numbers

$$0,1,2,\dots ,$$

where $n$$n$ is interpreted as ${\mathbb{R}}^n$$\mathbb R^n$.

The morphisms are

$${\mathcal{L}}_{C^{\mathrm{\infty }}}(n,m)=C^{\mathrm{\infty }}({\mathbb{R}}^n,{\mathbb{R}}^m).$$

Composition is ordinary composition of smooth maps.

The object $n$$n$ is the $n$$n$-fold product of $1$$1$, because for every $k$$k$ there is a natural bijection

$${\mathcal{L}}_{C^{\mathrm{\infty }}}(k,n)=C^{\mathrm{\infty }}({\mathbb{R}}^k,{\mathbb{R}}^n)\cong C^{\mathrm{\infty }}({\mathbb{R}}^k,\mathbb{R}{)}^n={\mathcal{L}}_{C^{\mathrm{\infty }}}(k,1{)}^n.$$

A $C^{\mathrm{\infty }}$$C^\infty$-ring is a finite-product-preserving functor

$$A:{\mathcal{L}}_{C^{\mathrm{\infty }}}\to \mathbf{Set}.$$

Its underlying set is

$$A(1).$$

Since $A$$A$ preserves finite products, one has

$$A(n)\cong A(1{)}^n.$$

Thus every smooth function

$$f:{\mathbb{R}}^n\to \mathbb{R}$$

induces an $n$$n$-ary operation

$$f_A:A^n\to A.$$

These operations satisfy exactly the equations imposed by projections and composition.

First, if

$${\pi }_i:{\mathbb{R}}^n\to \mathbb{R}$$

is the $i$$i$-th projection, then

$$({\pi }_i{)}_A(a_1,\dots ,a_n)=a_i.$$

Second, if

$$f_1,\dots ,f_m:{\mathbb{R}}^n\to \mathbb{R}$$

and

$$g:{\mathbb{R}}^m\to \mathbb{R},$$

then

$$(g\circ (f_1,\dots ,f_m){)}_A=g_A((f_1{)}_A,\dots ,(f_m{)}_A).$$

So a $C^{\mathrm{\infty }}$$C^\infty$-ring is not merely an $\mathbb{R}$$\mathbb R$-algebra with extra decoration. It is a model of a whole algebraic theory whose operations are all smooth functions.

Smooth manifolds give $C^{\mathrm{\infty }}$$C^\infty$-rings

Let $M$$M$ be a finite-dimensional Hausdorff second-countable smooth manifold.

Then

$$C^{\mathrm{\infty }}(M)$$

is naturally a $C^{\mathrm{\infty }}$$C^\infty$-ring.

Given

$$F\in C^{\mathrm{\infty }}({\mathbb{R}}^n,\mathbb{R})$$

and

$$g_1,\dots ,g_n\in C^{\mathrm{\infty }}(M),$$

define

$$F_{C^{\mathrm{\infty }}(M)}(g_1,\dots ,g_n)=F\circ (g_1,\dots ,g_n).$$

Explicitly,

$$p\mapsto F(g_1(p),\dots ,g_n(p)).$$

This is again a smooth function on $M$$M$.

A smooth map

$$\phi :M\to N$$

induces a pullback morphism

$${\phi }^{\ast }:C^{\mathrm{\infty }}(N)\to C^{\mathrm{\infty }}(M),f\mapsto f\circ \phi .$$

This pullback preserves all smooth operations, hence is a morphism of $C^{\mathrm{\infty }}$$C^\infty$-rings.

Therefore one obtains a contravariant functor

$$C^{\mathrm{\infty }}(-):{\mathbf{Man}}^{op}\to {\mathbf{C}}^{\mathrm{\infty }}\mathbf{Ring}.$$

For ordinary smooth manifolds, this functor is fully faithful:

$${\mathrm{Hom}}_{\mathbf{Man}}(M,N)\cong {\mathrm{Hom}}_{{\mathbf{C}}^{\mathrm{\infty }}\mathbf{Ring}}(C^{\mathrm{\infty }}(N),C^{\mathrm{\infty }}(M)).$$

This is the smooth analogue of the algebraic-geometric principle that functions remember the space.

Here is the idea.

Suppose

$$\mathrm{\Phi }:C^{\mathrm{\infty }}(N)\to C^{\mathrm{\infty }}(M)$$

is a morphism of $C^{\mathrm{\infty }}$$C^\infty$-rings.

For every point $x\in M$$x\in M$, compose with evaluation at $x$$x$:

$$C^{\mathrm{\infty }}(N)\stackrel{\mathrm{\Phi }}{\to }{C}^{\mathrm{\infty }}(M)\stackrel{{\mathrm{ev}}_x}{\to }\mathbb{R}.$$

This gives a $C^{\mathrm{\infty }}$$C^\infty$-ring morphism

$$C^{\mathrm{\infty }}(N)\to \mathbb{R}.$$

But the $\mathbb{R}$$\mathbb R$-points of $C^{\mathrm{\infty }}(N)$$C^\infty(N)$ are exactly the points of $N$$N$. Thus there is a unique point $\phi (x)\in N$$\varphi(x)\in N$ such that

$${\mathrm{ev}}_x\circ \mathrm{\Phi }={\mathrm{ev}}_{\phi (x)}.$$

So $\mathrm{\Phi }$$\Phi$ defines a set-map

$$\phi :M\to N.$$

For every $f\in C^{\mathrm{\infty }}(N)$$f\in C^\infty(N)$, one has

$$\mathrm{\Phi }(f)(x)=f(\phi (x)).$$

Hence

$$\mathrm{\Phi }(f)=f\circ \phi .$$

The fact that $\mathrm{\Phi }(f)$$\Phi(f)$ is smooth for every smooth $f$$f$ implies that $\phi$$\varphi$ is smooth. Locally, this follows by testing against coordinate functions, extended by bump functions if necessary.

Thus

$$\mathrm{\Phi }={\phi }^{\ast }.$$

So smooth maps are exactly $C^{\mathrm{\infty }}$$C^\infty$-ring morphisms in the opposite direction.

The smooth functor of points

Once manifolds are represented by their $C^{\mathrm{\infty }}$$C^\infty$-rings of functions, we may define the functor of points of a manifold $M$$M$ by

$$M(A)={\mathrm{Hom}}_{{\mathbf{C}}^{\mathrm{\infty }}\mathbf{Ring}}(C^{\mathrm{\infty }}(M),A),$$

where $A$$A$ is any $C^{\mathrm{\infty }}$$C^\infty$-ring.

This gives a functor

$$M(-):{\mathbf{C}}^{\mathrm{\infty }}\mathbf{Ring}\to \mathbf{Set}.$$

When $A=\mathbb{R}$$A=\mathbb R$,

$$M(\mathbb{R})={\mathrm{Hom}}_{{\mathbf{C}}^{\mathrm{\infty }}\mathbf{Ring}}(C^{\mathrm{\infty }}(M),\mathbb{R})\cong M.$$

Thus $\mathbb{R}$$\mathbb R$-points recover the ordinary points of $M$$M$.

But the real advantage of this language is that we can test $M$$M$ against infinitesimal $C^{\mathrm{\infty }}$$C^\infty$-rings, such as the dual numbers

$$D=\mathbb{R}[\epsilon ]/({\epsilon }^2).$$

This object is not the function ring of an honest manifold. It has a nonzero nilpotent element:

$$\epsilon \ne 0,{\epsilon }^2=0.$$

By contrast, $C^{\mathrm{\infty }}(M)$$C^\infty(M)$ is reduced: if $f^n=0$$f^n=0$, then $f=0$$f=0$ pointwise.

Therefore ordinary manifolds alone do not contain enough test objects to see infinitesimal directions. To see tangent vectors functorially, we must enlarge the test category from manifolds to $C^{\mathrm{\infty }}$$C^\infty$-rings.

The dual numbers as a $C^{\mathrm{\infty }}$$C^\infty$-ring

The algebra

$$D=\mathbb{R}[\epsilon ]/({\epsilon }^2)$$

has a natural $C^{\mathrm{\infty }}$$C^\infty$-ring structure.

Given a smooth function

$$F:{\mathbb{R}}^n\to \mathbb{R},$$

define

$$F_D:D^n\to D$$

by the first-order Taylor formula:

$$F_D(a_1+b_1\epsilon ,\dots ,a_n+b_n\epsilon )=F(a_1,\dots ,a_n)+\epsilon \sum _i{b}_i\frac{\partial F}{\partial x_i}(a_1,\dots ,a_n).$$

This is well-defined because ${\epsilon }^2=0$$\varepsilon^2=0$, so all higher-order terms vanish.

The compatibility with composition is exactly the chain rule. If

$$g:{\mathbb{R}}^m\to \mathbb{R}$$

and

$$f_1,\dots ,f_m:{\mathbb{R}}^n\to \mathbb{R},$$

then

$$d(g\circ (f_1,\dots ,f_m))=\sum _j\frac{\partial g}{\partial y_j}(f_1,\dots ,f_m)d{f}_j.$$

This is precisely the Lawvere-theoretic composition rule for the induced operations on $D$$D$.

Therefore $D$$D$ is a genuine $C^{\mathrm{\infty }}$$C^\infty$-ring.

Tangent vectors as dual-number points

Let

$$p:D\to \mathbb{R}$$

be the projection

$$a+b\epsilon \mapsto a.$$

It induces a map

$$M(D)\to M(\mathbb{R})\cong M.$$

For a point $x\in M$$x\in M$, define

$$T_{x}M=\{\varphi :C^{\mathrm{\infty }}(M)\to D|p\circ \varphi ={\mathrm{ev}}_x\}.$$

Here $\varphi$$\phi$ is required to be a $C^{\mathrm{\infty }}$$C^\infty$-ring morphism.

This is the smooth analogue of the algebraic-geometric definition of tangent vectors by dual numbers.

We now prove that this definition recovers the usual algebraic description of the tangent space as point derivations.

The key verification: dual-number points are point derivations

Fix $x\in M$$x\in M$. We want to prove the following proposition.

Proposition. There is a natural bijection

$$\{\varphi :C^{\mathrm{\infty }}(M)\to D|p\circ \varphi ={\mathrm{ev}}_x\}\cong {\mathrm{Der}}_{{\mathrm{ev}}_x}(C^{\mathrm{\infty }}(M),\mathbb{R}),$$

where the right-hand side is the set of $\mathbb{R}$$\mathbb R$-linear maps

$$X:C^{\mathrm{\infty }}(M)\to \mathbb{R}$$

satisfying the Leibniz rule at $x$$x$:

$$X(fg)=f(x)X(g)+g(x)X(f).$$

Thus a $D$$D$-point of $M$$M$ over $x$$x$ is the same thing as a tangent vector at $x$$x$.

Why the formula has this shape

Let

$$\varphi :C^{\mathrm{\infty }}(M)\to D$$

be a $D$$D$-point lying over $x$$x$. The condition

$$p\circ \varphi ={\mathrm{ev}}_x$$

means that for every smooth function $f\in C^{\mathrm{\infty }}(M)$$f\in C^\infty(M)$, the real part of $\varphi (f)$$\phi(f)$ is $f(x)$$f(x)$.

Since every element of $D$$D$ is uniquely of the form

$$a+b\epsilon ,$$

there is a unique real number $X(f)$$X(f)$ such that

$$\varphi (f)=f(x)+\epsilon X(f).$$

So any such $\varphi$$\phi$ automatically determines a function

$$X:C^{\mathrm{\infty }}(M)\to \mathbb{R}.$$

Because $\varphi$$\phi$ is multiplicative,

$$\varphi (fg)=\varphi (f)\varphi (g).$$

Using

$$\varphi (f)=f(x)+\epsilon X(f),\varphi (g)=g(x)+\epsilon X(g),$$

we get

$$f(x)g(x)+\epsilon X(fg)=(f(x)+\epsilon X(f))(g(x)+\epsilon X(g)).$$

Since ${\epsilon }^2=0$$\varepsilon^2=0$, the right-hand side is

$$f(x)g(x)+\epsilon (f(x)X(g)+g(x)X(f)).$$

Comparing the coefficient of $\epsilon$$\varepsilon$, we obtain

$$X(fg)=f(x)X(g)+g(x)X(f).$$

Thus $X$$X$ is a derivation at $x$$x$.

This proves one direction:

$$D\text{-point over }x⟹\text{point derivation at }x.$$

The converse is subtler.

Suppose we start with a point derivation

$$X:C^{\mathrm{\infty }}(M)\to \mathbb{R}.$$

The only possible candidate for the corresponding $D$$D$-point is

$${\varphi }_X(f)=f(x)+\epsilon X(f).$$

The Leibniz rule shows that ${\varphi }_X$$\phi_X$ is an ordinary $\mathbb{R}$$\mathbb R$-algebra morphism. But a $C^{\mathrm{\infty }}$$C^\infty$-ring morphism must preserve more than addition and multiplication. It must preserve every smooth operation.

Therefore we must prove that for every smooth function

$$F:{\mathbb{R}}^n\to \mathbb{R}$$

and every

$$f_1,\dots ,f_n\in C^{\mathrm{\infty }}(M),$$

one has

$${\varphi }_X(F(f_1,\dots ,f_n))=F_D({\varphi }_X(f_1),\dots ,{\varphi }_X(f_n)).$$

This is the real content of the proof.

The $C^{\mathrm{\infty }}$$C^\infty$-operation on the dual numbers

Recall the $C^{\mathrm{\infty }}$$C^\infty$-ring structure on

$$D=\mathbb{R}[\epsilon ]/({\epsilon }^2).$$

For a smooth function

$$F:{\mathbb{R}}^n\to \mathbb{R},$$

the induced operation

$$F_D:D^n\to D$$

is defined by first-order Taylor expansion:

$$F_D(a_1+b_1\epsilon ,\dots ,a_n+b_n\epsilon )=F(a_1,\dots ,a_n)+\epsilon \sum _i{b}_i\frac{\partial F}{\partial y_i}(a_1,\dots ,a_n).$$

Now apply this to

$${\varphi }_X(f_i)=f_i(x)+\epsilon X(f_i).$$

Then

$$F_D({\varphi }_X(f_1),\dots ,{\varphi }_X(f_n))$$

equals

$$F(f_1(x),\dots ,f_n(x))+\epsilon \sum _i\frac{\partial F}{\partial y_i}(f_1(x),\dots ,f_n(x))X(f_i).$$

On the other hand,

$${\varphi }_X(F(f_1,\dots ,f_n))=F(f_1,\dots ,f_n)(x)+\epsilon X(F(f_1,\dots ,f_n)).$$

Since

$$F(f_1,\dots ,f_n)(x)=F(f_1(x),\dots ,f_n(x)),$$

the constant terms already agree.

Thus the desired equality is equivalent to the chain-rule identity

$$X(F(f_1,\dots ,f_n))=\sum _i\frac{\partial F}{\partial y_i}(f_1(x),\dots ,f_n(x))X(f_i).$$

So the proof reduces to showing:

$$\boxed{{\displaystyle \text{A point derivation satisfying Leibniz automatically satisfies the chain rule.}}}$$

This is where the Hadamard lemma enters.

Lemma: point derivations kill constants

Let

$$X:C^{\mathrm{\infty }}(M)\to \mathbb{R}$$

be an ${\mathrm{ev}}_x$$\operatorname{ev}_x$-derivation. Then $X$$X$ kills constant functions.

Indeed,

$$X(1)=X(1\cdot 1)=1\cdot X(1)+1\cdot X(1)=2X(1).$$

Hence

$$X(1)=0.$$

For a constant $c\in \mathbb{R}$$c\in\mathbb R$,

$$X(c)=X(c\cdot 1)=cX(1)=0.$$

Thus

$$X(c)=0$$

for every constant function $c$$c$.

Lemma: Hadamard lemma

Let

$$F:{\mathbb{R}}^n\to \mathbb{R}$$

be smooth, and fix a point

$$a=(a_1,\dots ,a_n)\in {\mathbb{R}}^n.$$

Then there exist smooth functions

$$H_1,\dots ,H_n:{\mathbb{R}}^n\to \mathbb{R}$$

such that

$$F(y)-F(a)=\sum _i(y_i-a_i)H_i(y),$$

and

$$H_i(a)=\frac{\partial F}{\partial y_i}(a).$$

Here is a proof.

For $y\in {\mathbb{R}}^n$$y\in\mathbb R^n$, write

$$F(y)-F(a)={\int }_0^1\frac{d}{dt}F(a+t(y-a))dt.$$

By the chain rule,

$$\frac{d}{dt}F(a+t(y-a))=\sum _i(y_i-a_i)\frac{\partial F}{\partial y_i}(a+t(y-a)).$$

Therefore

$$F(y)-F(a)=\sum _i(y_i-a_i){\int }_0^1\frac{\partial F}{\partial y_i}(a+t(y-a))dt.$$

Define

$$H_i(y)={\int }_0^1\frac{\partial F}{\partial y_i}(a+t(y-a))dt.$$

Then each $H_i$$H_i$ is smooth, and

$$H_i(a)={\int }_0^1\frac{\partial F}{\partial y_i}(a)dt=\frac{\partial F}{\partial y_i}(a).$$

This proves the lemma.

Lemma: point derivations satisfy the chain rule

Let

$$X:C^{\mathrm{\infty }}(M)\to \mathbb{R}$$

be an ${\mathrm{ev}}_x$$\operatorname{ev}_x$-derivation. Let

$$F:{\mathbb{R}}^n\to \mathbb{R}$$

be smooth, and let

$$f_1,\dots ,f_n\in C^{\mathrm{\infty }}(M).$$

Set

$$a=(a_1,\dots ,a_n)=(f_1(x),\dots ,f_n(x)).$$

We claim that

$$X(F(f_1,\dots ,f_n))=\sum _i\frac{\partial F}{\partial y_i}(a)X(f_i).$$

By the Hadamard lemma, there exist smooth functions $H_i$$H_i$ such that

$$F(y)-F(a)=\sum _i(y_i-a_i)H_i(y),$$

with

$$H_i(a)=\frac{\partial F}{\partial y_i}(a).$$

Substitute

$$y_i=f_i.$$

Then in $C^{\mathrm{\infty }}(M)$$C^\infty(M)$,

$$F(f_1,\dots ,f_n)-F(a)=\sum _i(f_i-a_i)H_i(f_1,\dots ,f_n).$$

Apply $X$$X$ to both sides. Since $F(a)$$F(a)$ is a constant, $X(F(a))=0$$X(F(a))=0$. Therefore

$$X(F(f_1,\dots ,f_n))=\sum _{i}X((f_i-a_i)H_i(f_1,\dots ,f_n)).$$

Using the Leibniz rule,

$$X((f_i-a_i)H_i(f_1,\dots ,f_n))=(f_i(x)-a_i)X(H_i(f_1,\dots ,f_n))+H_i(f_1(x),\dots ,f_n(x))X(f_i-a_i).$$

But

$$f_i(x)-a_i=0,$$

so the first term vanishes.

Also $a_i$$a_i$ is constant, hence

$$X(f_i-a_i)=X(f_i).$$

Therefore

$$X((f_i-a_i)H_i(f_1,\dots ,f_n))=H_i(a)X(f_i).$$

Summing over $i$$i$ gives

$$X(F(f_1,\dots ,f_n))=\sum _i{H}_i(a)X(f_i).$$

Since

$$H_i(a)=\frac{\partial F}{\partial y_i}(a),$$

we get

$$X(F(f_1,\dots ,f_n))=\sum _i\frac{\partial F}{\partial y_i}(a)X(f_i).$$

This proves the chain rule for point derivations.

Completion of the proof

Now suppose

$$X:C^{\mathrm{\infty }}(M)\to \mathbb{R}$$

is an ${\mathrm{ev}}_x$$\operatorname{ev}_x$-derivation.

Define

$${\varphi }_X:C^{\mathrm{\infty }}(M)\to D$$

by

$${\varphi }_X(f)=f(x)+\epsilon X(f).$$

The Leibniz rule shows that ${\varphi }_X$$\phi_X$ preserves multiplication. Linearity shows that it preserves addition. Since $X(1)=0$$X(1)=0$, it preserves the unit. Since $X$$X$ kills constants, it is a morphism of $\mathbb{R}$$\mathbb R$-algebras.

It remains only to show that it preserves all smooth operations.

Let

$$F:{\mathbb{R}}^n\to \mathbb{R}$$

be smooth and let

$$f_1,\dots ,f_n\in C^{\mathrm{\infty }}(M).$$

The left-hand side is

$${\varphi }_X(F(f_1,\dots ,f_n))=F(f_1(x),\dots ,f_n(x))+\epsilon X(F(f_1,\dots ,f_n)).$$

By the chain-rule lemma,

$$X(F(f_1,\dots ,f_n))=\sum _i\frac{\partial F}{\partial y_i}(f_1(x),\dots ,f_n(x))X(f_i).$$

Therefore

$${\varphi }_X(F(f_1,\dots ,f_n))$$

equals

$$F(f_1(x),\dots ,f_n(x))+\epsilon \sum _i\frac{\partial F}{\partial y_i}(f_1(x),\dots ,f_n(x))X(f_i).$$

But this is exactly

$$F_D({\varphi }_X(f_1),\dots ,{\varphi }_X(f_n)).$$

Hence

$${\varphi }_X(F(f_1,\dots ,f_n))=F_D({\varphi }_X(f_1),\dots ,{\varphi }_X(f_n)).$$

So ${\varphi }_X$$\phi_X$ is a morphism of $C^{\mathrm{\infty }}$$C^\infty$-rings.

We have now proved both directions:

$$\varphi :C^{\mathrm{\infty }}(M)\to D\text{ over }x⟺X:C^{\mathrm{\infty }}(M)\to \mathbb{R}\text{ an }{\mathrm{ev}}_x\text{-derivation}.$$

Therefore

$$\{\varphi :C^{\mathrm{\infty }}(M)\to D|p\circ \varphi ={\mathrm{ev}}_x\}\cong {\mathrm{Der}}_{{\mathrm{ev}}_x}(C^{\mathrm{\infty }}(M),\mathbb{R}).$$

This is the algebraic definition of the tangent space:

$$T_{x}M\cong {\mathrm{Der}}_{{\mathrm{ev}}_x}(C^{\mathrm{\infty }}(M),\mathbb{R}).$$

Equivalently, using germs,

$$T_{x}M\cong {\mathrm{Der}}_{\mathbb{R}}(C_{M,x}^{\mathrm{\infty }},\mathbb{R}).$$

Thus the tangent space is exactly the fiber over $x$$x$ of

$$M(D)\to M(\mathbb{R}).$$

In this sense, tangent vectors are dual-number points.

The total tangent bundle as the set of dual-number points

The previous section described the fiber of

$$M(D)\to M(\mathbb{R})$$

over a point $x\in M$$x\in M$ as the tangent space $T_{x}M$$T_xM$.

It is worth saying the global version explicitly:

$$\boxed{{\displaystyle M(D)\cong TM.}}$$

Here

$$D=\mathbb{R}[\epsilon ]/({\epsilon }^2)$$

is regarded as a $C^{\mathrm{\infty }}$$C^\infty$-ring, and

$$M(D)={\mathrm{Hom}}_{C^{\mathrm{\infty }}\mathrm{Ring}}(C^{\mathrm{\infty }}(M),D).$$

If one defines $C^{\mathrm{\infty }}$$C^\infty$-rings as models of the Lawvere theory ${\mathcal{L}}_{C^{\mathrm{\infty }}}$$\mathcal L_{C^\infty}$, then this Hom-set can also be written as

$${\mathrm{Nat}}_{{\mathcal{L}}_{C^{\mathrm{\infty }}}}(C^{\mathrm{\infty }}(M),D),$$

because morphisms of $C^{\mathrm{\infty }}$$C^\infty$-rings are exactly natural transformations between the corresponding finite-product-preserving functors

$${\mathcal{L}}_{C^{\mathrm{\infty }}}\to \mathbf{Set}.$$

Thus the set of dual-number points of $M$$M$ is

$$M(D)={\mathrm{Nat}}_{{\mathcal{L}}_{C^{\mathrm{\infty }}}}(C^{\mathrm{\infty }}(M),\mathbb{R}[\epsilon ]/{\epsilon }^2).$$

The projection

$$p:D\to \mathbb{R},a+b\epsilon \mapsto a,$$

induces

$$M(D)\to M(\mathbb{R})\cong M.$$

The fiber over $x\in M$$x\in M$ is

$${(M(D)\to M(\mathbb{R}))}^{-1}(x)\cong T_{x}M.$$

Therefore

$$M(D)\cong \coprod _{x\in M}{T}_{x}M.$$

This is the underlying set of the tangent bundle.

The smooth manifold structure on $M(D)$$M(D)$ is also the usual smooth structure on $TM$$TM$. To see this, take a coordinate chart

$$U\subseteq {\mathbb{R}}^n.$$

A $D$$D$-point of $U$$U$ is an $n$$n$-tuple

$$(a_1+b_1\epsilon ,\dots ,a_n+b_n\epsilon )\in D^n$$

whose real part

$$(a_1,\dots ,a_n)$$

lies in $U$$U$. Thus

$$U(D)\cong U\times {\mathbb{R}}^n.$$

The first component is the base point, and the second component is the tangent vector.

Now let

$$\psi :U\to V$$

be a smooth coordinate change. Its extension to $D$$D$-points is given by first-order Taylor expansion:

$${\psi }_D(a+b\epsilon )=\psi (a)+\epsilon d{\psi }_a(b).$$

Hence the transition function between local descriptions

$$U(D)\cong U\times {\mathbb{R}}^n$$

and

$$V(D)\cong V\times {\mathbb{R}}^n$$

is

$$(a,b)\mapsto (\psi (a),d{\psi }_a(b)),$$

which is exactly the usual transition function for the tangent bundle.

So the functorial construction does not merely recover each tangent space separately. It recovers the entire tangent bundle:

$$\boxed{{\displaystyle TM\cong M(\mathbb{R}[\epsilon ]/{\epsilon }^2).}}$$

Under this identification, the tangent bundle projection

$$TM\to M$$

is induced by the $C^{\mathrm{\infty }}$$C^\infty$-ring morphism

$$D\to \mathbb{R}.$$

The zero section

$$M\to TM$$

is induced by the inclusion

$$\mathbb{R}\to D,a\mapsto a.$$

Fiberwise scalar multiplication by $\lambda \in \mathbb{R}$$\lambda\in\mathbb R$ is induced by

$$D\to D,\epsilon \mapsto \lambda \epsilon .$$

Fiberwise addition is induced by the morphism

$$D{\times }_{\mathbb{R}}D\to D,a+b{\epsilon }_1+c{\epsilon }_2\mapsto a+(b+c)\epsilon .$$

Here

$$D{\times }_{\mathbb{R}}D\cong \mathbb{R}[{\epsilon }_1,{\epsilon }_2]/({\epsilon }_1^2,{\epsilon }_2^2,{\epsilon }_1{\epsilon }_2).$$

Thus even the vector bundle structure of $TM$$TM$ is encoded by the elementary algebra of dual numbers.

This is the precise smooth analogue of the algebraic-geometric slogan:

$$\text{tangent vectors are maps from the dual numbers.}$$