nLab Tietze extension theorem

Contents

Context

Differential geometry

synthetic differential geometry

Introductions

from point-set topology to differentiable manifolds

Differentials

V-manifolds

smooth space

Tangency

The magic algebraic facts

Theorems

Axiomatics

cohesion

tangent cohesion

differential cohesion

$\array{ && id &\dashv& id \\ && \vee && \vee \\ &\stackrel{fermionic}{}& \rightrightarrows &\dashv& \rightsquigarrow & \stackrel{bosonic}{} \\ && \bot && \bot \\ &\stackrel{bosonic}{} & \rightsquigarrow &\dashv& \mathrm{R}\!\!\mathrm{h} & \stackrel{rheonomic}{} \\ && \vee && \vee \\ &\stackrel{reduced}{} & \Re &\dashv& \Im & \stackrel{infinitesimal}{} \\ && \bot && \bot \\ &\stackrel{infinitesimal}{}& \Im &\dashv& \& & \stackrel{\text{étale}}{} \\ && \vee && \vee \\ &\stackrel{cohesive}{}& ʃ &\dashv& \flat & \stackrel{discrete}{} \\ && \bot && \bot \\ &\stackrel{discrete}{}& \flat &\dashv& \sharp & \stackrel{continuous}{} \\ && \vee && \vee \\ && \emptyset &\dashv& \ast }$

Models

Lie theory, ∞-Lie theory

differential equations, variational calculus

Chern-Weil theory, ∞-Chern-Weil theory

Cartan geometry (super, higher)

Contents

Summary

The Tietze extension theorem says that continuous functions extend from closed subsets of a normal topological space $X$ to the whole space $X$.

This is a close cousin of Urysohn's lemma with many applications.

One implication is that topological vector bundles over a topological space $X$ that trivialize over a closed subspace $A$ are equivalent to vector bundles on the quotient space $X/A$ (see there). This in turn is what implies the long exact sequence in cohomology for topological K-theory (see there).

Statement

For continuous functions

Theorem

For $X$ a normal topological space and $A \subset X$ a closed subspace, there is for every continuous function $f \colon A \to \mathbb{R}$ to the real line (with its Euclidean metric topology) a continuous function $\hat f \colon X \to \mathbb{R}$ extending it, i.e. such that $\hat f|_A = f$:

$\array{ A &\hookrightarrow & X \\ {}^{\mathllap{f}}\downarrow & \swarrow_{\mathrlap{\exists \hat f} } \\ \mathbb{R} }$

Therefore one also says that $\mathbb{R}$ is an absolute extensor in topology.

Proof

We produce a sequence of approximations to the desired extension by induction. Then we will observe that the sequence is a Cauchy sequence and conclude by observing that this implies that it limit is an extension of $f$ as desired.

For the induction step, let

$\hat f_n \;\colon\; X \longrightarrow \mathbb{R}$

be a continuous function on $X$ such that the difference of its restriction to $A$ with $f$ is a bounded function, for a bound $c_n \in (0,\infty) \subset \mathbb{R}$:

$\underset{a \in A}{\forall} \left( { \left\Vert f(a) - \hat f_n (a) \right\Vert } \leq c_n \right) \,.$

Consider then the pre-image subsets

$S_- \coloneqq \left( f - \hat f_n\vert_A \right)^{-1}\big( [-c_n, -c_n/3] \big) \phantom{AAAA} S_+ \coloneqq \left( f - \hat f_n\vert_A \right)^{-1}\big( [c_n/3, c_n] \big) \,.$

Since the closed intervals $[-c_n,-c_n/3], [c_n/3, c_n] \subset \mathbb{R}$ are closed subsets, and since $f - \hat f_n\vert_A$ is a continuous function, these are closed subsets of $A$. Moreover, since subsets are closed in a closed subspace precisely if they are closed in the ambient space, these are also closed subsets of $X$.

Therefore, since $X$ is normal by assumption, it follows with Urysohn's lemma that there is a continuous function

$\phi \;\colon\; X \longrightarrow \mathbb{R}$

with

$\underset{x \in X}{\forall} \left( 0 \leq \phi(x) \leq 1 \right)$

and

$\phi\vert_{S_+} = 1 \phantom{AAAA} \phi\vert_{S_-} = 0 \,.$

Consider then the continuous function

$g_{n+1} \;\coloneqq\; \tfrac{2 c_n}{3} \phi - \tfrac{c_n}{3}$

This now satisfies

$g_{n+1}\vert_{S_+} = \frac{c_n}{3} \phantom{AAAA} g_{n+1}\vert_{S_-} = -\frac{c_n}{3} \,.$

with

$\underset{x \in X}{\forall} \left( \left \Vert g_{n+1} (x) \right\Vert \leq \tfrac{c_n}{3} \right) \,.$

Moreover, observe that this function satisfies

$\underset{a \in A}{\forall} \left( \left\Vert f - \hat f_n(a) - g_{n+1}(a) \right\Vert \leq \tfrac{2 c_n}{3} \right) \,.$

To wit, this is because

1. for $a \in S_+$ we have $g_{n+1}(a) = \tfrac{c_n}{3}$ and $f(a) - \hat f_{n}(a) \in [c_n/3,c_n]$;

2. for $a \in S_-$ we have $g_{n+1}(a) = -\tfrac{c_n}{3}$ and $f(a) - \hat f_{n}(a) \in [-c_n/3,-c_n]$;

3. for $a \in Y \setminus \{S_+ \cup S_-\}$ we have $g(a) \in [-c_n/3,c_n/3]$ as well as $f(a) - \hat f_{n}(a) \in [-c_n/3, c_n/3]$.

It follows that if we set

$\hat f_{n+1} \coloneqq \hat f_n + g_{n+1}$

then

$\underset{a \in A}{\forall} \left( { \left\Vert f(a) - \hat f_{n+1}(a) \right\Vert } \leq \tfrac{2 c_n}{3} \right) \,.$

This gives the induction step.

To start the induction, first assume that $f$ is bounded by a constant $c_0$. Then we may set

$\hat f_0 \coloneqq const_0 \,.$

Hence induction now gives a sequence of continuous functions

$(\hat f_n)_{n \in \mathbb{N}}$

with the property that

$\underset{a \in A}{\forall} \left( \left\Vert f(a) -\hat f_n(a) \right\Vert \leq \left( \tfrac{2}{3}\right)^n c_0 \right) \,.$

Moreover, for $n_1, n_2 \in \mathbb{N}$ with $n_2 \geq n_1$ and $x \in X$ we have

\begin{aligned} {\Vert \hat f_{n_2}(x) - \hat f_{n_1}(x) \Vert} & = {\Vert g_{n_1 + 1}(x) + g_{n_1 + 2}(c) + \cdots + g_{n_2}(x) \Vert} \\ & \leq \underoverset{k = n_1+1}{n_2}{\sum} \tfrac{1}{3^{k}} c_0 \\ & \leq \underoverset{k = n_1+1}{\infty}{\sum} \tfrac{1}{3^{k}} c_0 \end{aligned}

That the geometric series $\sum_{k = 0}^\infty 1/3^k$ converges

$\underoverset{n}{k = 0}{\sum} 1/3 k \overset{n \to \infty}{\longrightarrow} \frac{1}{1 - 1/3} = 3/2$

this becomes arbitrarily small for large $n_1$.

This means that the sequence $(\hat f_{n+1})_{n\in \mathbb{N}}$ is a Cauchy sequence in the supremum norm for real-valued functions.

Since uniform Cauchy sequences of continuous functions with values in a complete metric space converge uniformly to a continuous function (this prop.) this implies that the sequence converges uniformly to a continuous function. By construction, this is an extension as required.

Finally consider the case that $f$ is not a bounded function. In this case consider any homeomorphism $\phi \colon \mathbb{R}^1 \overset{\simeq}{\to} (-c_0,c_0) \subset \mathbb{R}^1$ between the real line and an open interval Then $\phi \circ f$ is a continous function bounded by $c_0$ and hence the above argument gives an extension $\widehat {\phi \circ f}$. Then $\phi^{-1} \circ \widehat{ \phi \circ f }$ is an extension of $f$.

For smooth loci

Let $\mathbb{L} = (C^\infty Ring^{fin})^{op}$ be the category of smooth loci, the opposite category of finitely generated generalized smooth algebras. By the theorem discussed there, there is a full and faithful functor Diff $\hookrightarrow \mathbb{L}$.

Definition

For $A = C^\infty(\mathbb{R}^n)/J$ and $B = C^\infty(\mathbb{R}^n)/I$ with $I \subset J$ and $B \to A$ the projection of generalized smooth algebras the corresponding monomorphism $\ell A \to \ell B$ in $\mathbb{L}$ exhibits $\ell A$ as a closed smooth sublocus of $\ell B$.

Lemma

Let $X$ be a smooth manifold and let $\{g_i \in C^\infty(X)\}_{i = 1}^n$ be smooth functions that are independent in the sense that at each common zero point $x\in X$, $\forall i : g_i(x)= 0$ we have the derivative $(d g_i) : T_x X \to \mathbb{R}^n$ is a surjection, then the ideal $(g_1, \cdots, g_n)$ coincides with the ideal of functions that vanish on the zero-set of the $g_i$.

This is lemma 2.1 in Chapter I of (MoerdijkReyes).

Proposition

If $\ell A \hookrightarrow \ell B$ is a closed sublocus of $\ell B$ then every morphism $\ell A \to R$ extends to a morphism $\ell B \to R$

This is prop. 1.6 in Chapter II of (MoerdijkReyes).

Proof

Since we have $R = \ell C^\infty(\mathbb{R})$ and $C^\infty(\mathbb{R})$ is the free generalized smooth algebra on a single generator, a morphism $\ell A \to R$ is precisely an element of $C^\infty(\mathbb{R}^n)/J$. This is represented by an element in $C^\infty(\mathbb{R}^n)$ which in particular defines an element in $C^\infty(\mathbb{R}^n)/I$.

Leture notes include

• Adam Boocher, A proof of the Tietze Extension Theorem Using Urysohn’s Lemma, 2005 (pdf)

Discussion of the smooth version includes