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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).
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 $\hat f|_A = f$:
Therefore one also says that $\mathbb{R}$ is an absolute extensor in topology.
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 its limit is an extension of $f$ as desired.
For the induction step, let
be a continuous function on $X$ such that the difference of its restriction to $A$ with $f$ is a bounded function, for a bound $c_n \in (0,\infty) \subset \mathbb{R}$:
Consider then the pre-image subsets
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
with
and
Consider then the continuous function
This now satisfies
with
Moreover, observe that this function satisfies
To wit, this is because
for $a \in S_+$ we have $g_{n+1}(a) = \tfrac{c_n}{3}$ and $f(a) - \hat f_{n}(a) \in [c_n/3,c_n]$;
for $a \in S_-$ we have $g_{n+1}(a) = -\tfrac{c_n}{3}$ and $f(a) - \hat f_{n}(a) \in [-c_n/3,-c_n]$;
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
then
This gives the induction step.
To start the induction, first assume that $f$ is bounded by a constant $c_0$. Then we may set
Hence induction now gives a sequence of continuous functions
with the property that
Moreover, for $n_1, n_2 \in \mathbb{N}$ with $n_2 \geq n_1$ and $x \in X$ we have
That the geometric series $\sum_{k = 0}^\infty 1/3^k$ converges
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 $f$.
See Whitney extension theorem, also Steenrod-Wockel approximation theorem.
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 $\hookrightarrow \mathbb{L}$.
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 $\ell B$.
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).
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).
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 $C^\infty(\mathbb{R}^n)/I$.
extension theorems | continuous functions | smooth functions |
---|---|---|
plain functions | Tietze extension theorem | Whitney extension theorem |
equivariant functions | equivariant Tietze extension theorem |
Named after:
Leture notes:
Generalization to maps into any locally convex topological vector space:
Discussion via algebraic topology:
Discussion of the smooth version includes
See also
Wikipedia, Tietze extension theorem
Bruce Blackadar, Extending continuous functions [arXiv:1207.6147]
Last revised on May 28, 2023 at 09:36:18. See the history of this page for a list of all contributions to it.