What is the utility of the Tietze extension theorem?

Question

In the notes I read, the Urysohn lemma is presented before the Tietze extension theorem. Urysohn lemma's utility is quite easy to understand. If we are given a topological space $X$, we want to be able to construct some functions on domain $X$.

I can appreciate this because in some general space , we may not have any additional algebraic structure like we do in $\mathbb{R}$, and hence it could be quite difficult to write down any such function.

But, then what is the use of having the Tietze extension theorem? Is there any natural setting where one would to extend functions on a closed subset of a topological space $X$ into the entire?

I don't know of any direct setting like you are asking, but there is a result in K-theory (namely that the induced sequence of [several] K-theory groups are exact) that uses the Tietze extension theorem for a crucial lemma used in the proof. Is this something you might be interested in? — HackR, Sep 21 '24 at 13:50
For something much much simpler, this answer sounds like a candidate https://math.stackexchange.com/a/122339/1088726 — HackR, Sep 21 '24 at 13:54

Martin Argerami · Accepted Answer · 2024-09-22T12:57:17.030

2

Here are two examples off the top of my head.

First occurs when proving that if you have an injection $\alpha:S\to T$ between two compact topological spaces, there exists a continuous surjective homomorphism $\pi_\alpha:C(T)\to C(S)$. The homomorphism is given by $\pi_\alpha(f)=f\circ\alpha$. Given $g\in C(S)$, one can define $f:\alpha(S)\to T$ by $f=g\circ \alpha^{-1}$. Extend this to $\tilde f\in C(T)$ by Tietze and then $\pi_\alpha(\tilde f)=g$.

Another use I know is when proving that the ideals in $C(T)$, for $T$ compact and Hausdorff, and precisely the subsets $J_K=\{f\in C(T):\ f|_K=0\}$.

edited Sep 22 '24 at 12:57

answered Sep 21 '24 at 23:21

Martin Argerami

217,281

1

What is C(T)..? – Clemens Bartholdy Sep 22 '24 at 06:23
I don't know if this theorem is correct. Do you have a reference for it? – Clemens Bartholdy Sep 22 '24 at 09:17
@CantorDustDrachen: $C(T)={f:T\to\mathbb C, \text{continuous}}$. As for the theorem, there's a proof of it in my answer, what do you see wrong with it? The result that compact Hausdorff spaces $S,T$ are homeomorphic if and only if $C(T)$ and $C(S)$ are isomorphic as algebras (or as rings) is often attributed to Gelfand-Kolmogorov. – Martin Argerami Sep 22 '24 at 12:55
Refer. It's quite insane that the ring structure of continous function determine the topological space entirely – Clemens Bartholdy Sep 22 '24 at 13:02
I guess this is the "right" answer – Clemens Bartholdy Sep 22 '24 at 13:04

score 2 · Answer 2 · answered Sep 22 '24 at 00:48

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The functional calculus for the spectral theorem for bounded, self-adjoint operators can be proven in a way that uses Tietze's extension theorem... to get from continuous functions on the spectrum to continuous functions on the real line. That kind of thing. It is mildly amazing to me, in that context, that any continuous function on a compact subset of $\mathbb R$ is a restriction of a continuous, compactly-supported function on $\mathbb R$. :)

answered Sep 22 '24 at 00:48

paul garrett

55,317

I guess this is also a good answer but unfortunately one can not answer multipile answers at once – Clemens Bartholdy Sep 22 '24 at 13:04

What is the utility of the Tietze extension theorem?

2 Answers2