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\begin{document}

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\section*{Tietze extension theorem}

\hypertarget{context}{}\subsubsection*{{Context}}\label{context}

\hypertarget{topology}{}\paragraph*{{Topology}}\label{topology}

[[!include topology - contents]]

\hypertarget{differential_geometry}{}\paragraph*{{Differential geometry}}\label{differential_geometry}

[[!include synthetic differential geometry - contents]]

\hypertarget{contents}{}\section*{{Contents}}\label{contents}

\noindent\hyperlink{summary}{Summary}\dotfill \pageref*{summary} \linebreak
\noindent\hyperlink{statement}{Statement}\dotfill \pageref*{statement} \linebreak
\noindent\hyperlink{ForContinuousFunctions}{For continuous functions}\dotfill \pageref*{ForContinuousFunctions} \linebreak
\noindent\hyperlink{Manifolds}{For smooth functions}\dotfill \pageref*{Manifolds} \linebreak
\noindent\hyperlink{for_smooth_loci}{For smooth loci}\dotfill \pageref*{for_smooth_loci} \linebreak
\noindent\hyperlink{related_theorems}{Related theorems}\dotfill \pageref*{related_theorems} \linebreak
\noindent\hyperlink{references}{References}\dotfill \pageref*{references} \linebreak


\hypertarget{summary}{}\subsection*{{Summary}}\label{summary}

The \emph{Tietze extension theorem} says that [[continuous functions]] [[extension|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 \href{topological+vector+bundle#OverClosedSubspaces}{there}). This in turn is what implies the [[long exact sequence in cohomology]] for [[topological K-theory]] (see \href{topological+K-theory#ExactSequences}{there}).

\hypertarget{statement}{}\subsection*{{Statement}}\label{statement}

\hypertarget{ForContinuousFunctions}{}\subsubsection*{{For continuous functions}}\label{ForContinuousFunctions}

\begin{theorem}
\label{}\hypertarget{}{}
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 space|Euclidean]] [[metric topology]]) a continuous function $\hat f \colon X \to \mathbb{R}$ [[extension|extending]] it, i.e. such that $\hat f|_A = f$:

\begin{displaymath}
\itexarray{
     A &\hookrightarrow & X
     \\
     {}^{\mathllap{f}}\downarrow & \swarrow_{\mathrlap{\exists \hat f} } 
     \\
     \mathbb{R}
  }
\end{displaymath}
Therefore one also says that \emph{$\mathbb{R}$ is an [[absolute extensor]]} in [[topology]].

\end{theorem}
\begin{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 of a sequence|limit]] is an extension of $f$ as desired.

For the induction step, let

\begin{displaymath}
\hat f_n \;\colon\; X \longrightarrow \mathbb{R}
\end{displaymath}
be a [[continuous function]] on $X$ such that the difference of its restriction to $A$ with $f$ is a [[bounded function]], for a bound $c_n \in (0,\infty) \subset \mathbb{R}$:

\begin{displaymath}
\underset{a \in A}{\forall} 
  \left( 
    { \left\Vert f(a) - \hat f_n (a) \right\Vert } \leq c_n
  \right)
  \,.
\end{displaymath}
Consider then the [[pre-image]] subsets

\begin{displaymath}
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)
  \,.
\end{displaymath}
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 topological space|normal]] by assumption, it follows with [[Urysohn's lemma]] that there is a continuous function

\begin{displaymath}
\phi \;\colon\; X \longrightarrow \mathbb{R}
\end{displaymath}
with

\begin{displaymath}
\underset{x \in X}{\forall} \left( 0 \leq \phi(x) \leq 1 \right)
\end{displaymath}
and

\begin{displaymath}
\phi\vert_{S_+} = 1
  \phantom{AAAA}
  \phi\vert_{S_-} = 0
  \,.
\end{displaymath}
Consider then the continuous function

\begin{displaymath}
g_{n+1}
    \;\coloneqq\;
  \tfrac{2 c_n}{3} \phi
    - 
  \tfrac{c_n}{3}
\end{displaymath}
This now satisfies

\begin{displaymath}
g_{n+1}\vert_{S_+} = \frac{c_n}{3}
  \phantom{AAAA}
  g_{n+1}\vert_{S_-} = -\frac{c_n}{3}
  \,.
\end{displaymath}
with

\begin{displaymath}
\underset{x \in X}{\forall} \left( 
    \left \Vert g_{n+1} (x) \right\Vert 
     \leq 
    \tfrac{c_n}{3}
  \right)
  \,.
\end{displaymath}
Moreover, observe that this function satisfies

\begin{displaymath}
\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)
 \,.
\end{displaymath}
To wit, this is because

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


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


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



\end{enumerate}
It follows that if we set

\begin{displaymath}
\hat f_{n+1} \coloneqq \hat f_n + g_{n+1}
\end{displaymath}
then

\begin{displaymath}
\underset{a \in A}{\forall}
  \left(
     { \left\Vert f(a) - \hat f_{n+1}(a) \right\Vert }
     \leq 
     \tfrac{2 c_n}{3}
  \right)
  \,.
\end{displaymath}
This gives the induction step.

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

\begin{displaymath}
\hat f_0 \coloneqq const_0
  \,.
\end{displaymath}
Hence [[induction]] now gives a [[sequence]] of continuous functions

\begin{displaymath}
(\hat f_n)_{n \in \mathbb{N}}
\end{displaymath}
with the property that

\begin{displaymath}
\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)
  \,.
\end{displaymath}
Moreover, for $n_1, n_2 \in \mathbb{N}$ with $n_2 \geq n_1$ and $x \in X$ we have

\begin{displaymath}
\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}
\end{displaymath}
That the [[geometric series]] $\sum_{k = 0}^\infty 1/3^k$ [[convergence of a sequence|converges]]

\begin{displaymath}
\underoverset{n}{k = 0}{\sum} 1/3 k
   \overset{n \to \infty}{\longrightarrow}
  \frac{1}{1 - 1/3} = 3/2
\end{displaymath}
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 space|complete]] [[metric space]] [[uniform convergence|converge uniformly]] to a [[continuous function]] (\href{uniform+convergence#FunctionsUniformCauchySequence}{this prop.}) this implies that the sequence [[uniform convergence|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 $f$.

\end{proof}
\hypertarget{Manifolds}{}\subsubsection*{{For smooth functions}}\label{Manifolds}

See \emph{[[Whitney extension theorem]]}, also \emph{[[Steenrod-Wockel approximation theorem]]}.

\hypertarget{for_smooth_loci}{}\subsubsection*{{For smooth loci}}\label{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 algebra]]s. By the theorem discussed there, there is a [[full and faithful functor]] [[Diff]] $\hookrightarrow \mathbb{L}$.

\begin{defn}
\label{}\hypertarget{}{}
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 algebra]]s the corresponding [[monomorphism]] $\ell A \to \ell B$ in $\mathbb{L}$ exhibits $\ell A$ as a \textbf{closed smooth sublocus} of $\ell B$.

\end{defn}
\begin{lemma}
\label{}\hypertarget{}{}
Let $X$ be a [[smooth manifold]] and let $\{g_i \in C^\infty(X)\}_{i = 1}^n$ be [[smooth function]]s 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 $g_i$.

\end{lemma}
This is lemma 2.1 in Chapter I of (\hyperlink{MoerdijkReyes}{MoerdijkReyes}).

\begin{prop}
\label{}\hypertarget{}{}
If $\ell A \hookrightarrow \ell B$ is a closed sublocus of $\ell B$ then every morphism $\ell A \to R$ extends to a morphism $\ell B \to R$

\end{prop}
This is prop. 1.6 in Chapter II of (\hyperlink{MoerdijkReyes}{MoerdijkReyes}).

\begin{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 $C^\infty(\mathbb{R}^n)/I$.

\end{proof}
\hypertarget{related_theorems}{}\subsection*{{Related theorems}}\label{related_theorems}

[[!include extension theorems -- table]]

\begin{itemize}%
\item [[Urysohn's lemma]]


\item [[Hadamard lemma]]


\item [[Borel's theorem]]


\item [[Steenrod-Wockel approximation theorem]]


\item [[Whitney extension theorem]]


\item [[Taimanov theorem]]


\item [[embedding of smooth manifolds into formal duals of R-algebras]]


\item [[smooth Serre-Swan theorem]]


\item [[derivations of smooth functions are vector fields]]



\end{itemize}
\hypertarget{references}{}\subsection*{{References}}\label{references}

Leture notes include

\begin{itemize}%
\item Adam Boocher, \emph{A proof of the Tietze Extension Theorem Using Urysohn's Lemma}, 2005 (\href{http://www.maths.ed.ac.uk/~aboocher/math/tietze.pdf}{pdf})

\end{itemize}
Discussion of the smooth version includes

\begin{itemize}%
\item [[Ieke Moerdijk]], [[Gonzalo Reyes]], chapters I and II of \emph{[[Models for Smooth Infinitesimal Analysis]]}

\end{itemize}
See also

\begin{itemize}%
\item Wikipedia, \emph{\href{https://en.wikipedia.org/wiki/Tietze_extension_theorem}{Tietze extension theorem}}


\item [[Planet Math]], \emph{\href{http://planetmath.org/proofoftietzeextensiontheorem}{Proof of the Tietze extension theorem}}



\end{itemize}


\end{document}