Tietze extension theorem



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

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

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


For continuous functions


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

A X f 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.


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 ff as desired.

For the induction step, let

f^ n:X \hat f_n \;\colon\; X \longrightarrow \mathbb{R}

be a continuous function on XX such that the difference of its restriction to AA with ff is a bounded function, for a bound c n(0,)c_n \in (0,\infty) \subset \mathbb{R}:

aA(f(a)f^ n(a)c n). \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 (ff^ n| A) 1([c n,c n/3])AAAAS +(ff^ n| A) 1([c n/3,c n]). 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][-c_n,-c_n/3], [c_n/3, c_n] \subset \mathbb{R} are closed subsets, and since ff^ n| Af - \hat f_n\vert_A is a continuous function, these are closed subsets of AA. Moreover, since subsets are closed in a closed subspace precisely if they are closed in the ambient space, these are also closed subsets of XX.

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

ϕ:X \phi \;\colon\; X \longrightarrow \mathbb{R}


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


ϕ| S +=1AAAAϕ| S =0. \phi\vert_{S_+} = 1 \phantom{AAAA} \phi\vert_{S_-} = 0 \,.

Consider then the continuous function

g n+12c n3ϕc n3 g_{n+1} \;\coloneqq\; \tfrac{2 c_n}{3} \phi - \tfrac{c_n}{3}

This now satisfies

g n+1| S +=c n3AAAAg n+1| S =c n3. g_{n+1}\vert_{S_+} = \frac{c_n}{3} \phantom{AAAA} g_{n+1}\vert_{S_-} = -\frac{c_n}{3} \,.


xX(g n+1(x)c n3). \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

aA(ff^ n(a)g n+1(a)2c n3). \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 aS +a \in S_+ we have g n+1(a)=c n3g_{n+1}(a) = \tfrac{c_n}{3} and f(a)f^ n(a)[c n/3,c n]f(a) - \hat f_{n}(a) \in [c_n/3,c_n];

  2. for aS a \in S_- we have g n+1(a)=c n3g_{n+1}(a) = -\tfrac{c_n}{3} and f(a)f^ n(a)[c n/3,c n]f(a) - \hat f_{n}(a) \in [-c_n/3,-c_n];

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

It follows that if we set

f^ n+1f^ n+g n+1 \hat f_{n+1} \coloneqq \hat f_n + g_{n+1}


aA(f(a)f^ n+1(a)2c n3). \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 ff is bounded by a constant c 0c_0. Then we may set

f^ 0const 0. \hat f_0 \coloneqq const_0 \,.

Hence induction now gives a sequence of continuous functions

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

with the property that

aA(f(a)f^ n(a)(23) nc 0). \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 2n_1, n_2 \in \mathbb{N} with n 2n 1n_2 \geq n_1 and xXx \in X we have

f^ n 2(x)f^ n 1(x) =g n 1+1(x)+g n 1+2(c)++g n 2(x) k=n 1+1n 213 kc 0 k=n 1+113 kc 0 \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 k=0 1/3 k\sum_{k = 0}^\infty 1/3^k converges

nk=01/3kn111/3=3/2 \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 1n_1.

This means that the sequence (f^ n+1) n(\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 ff is not a bounded function. In this case consider any homeomorphism ϕ: 1(c 0,c 0) 1\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 ϕf\phi \circ f is a continous function bounded by c 0c_0 and hence the above argument gives an extension ϕf^\widehat {\phi \circ f}. Then ϕ 1ϕf^\phi^{-1} \circ \widehat{ \phi \circ f } is an extension of ff.

For smooth functions

See Whitney extension theorem, also Steenrod-Wockel approximation theorem.

For smooth loci

Let 𝕃=(C Ring fin) op\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 𝕃\hookrightarrow \mathbb{L}.


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


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

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


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

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


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

See also

Revised on July 19, 2017 08:08:53 by Urs Schreiber (