# How does this imply surjectivity? An application of the Tietze extension theorem.

Consider the functor $C$ from the category of compact Hausdorff spaces to that of unital abelian C*-algebras given by $C: X \mapsto C(X, \mathbb{C})$ (the continuous maps from $X$ to $\mathbb{C}$) and also, for continuous $f: X \to Y$ ($Y$ being another compact Hausdorff space), $Cf: C(Y) \to C(X), h \mapsto h \circ f$. If $Cf$ is injective, it is then the case that $f$ is surjective -- my question is how/why? Several resources that I have found brush this off by merely invoking Tietze - and then moving on. However, I don't see how this works, exactly. Clearly I'm overlooking something elementary. Here is as far as I get with my reasoning.

First off, $X$ is a compact Hausdorff space and is hence normal. Moreover, $f(X)$ is a closed (and compact) subset of $Y$. If $Cf$ is injective, then we have that for $h_1,h_2 \in C(Y)$, if $h_1|_{f(X)} = h_2|_{f(X)}$ then it must be that $h_1 = h_2$.

I suppose a slight variant of Tietze's theorem is being considered, where in the real and imaginary parts of the function are being extended, however, I see no value in extending $h_1|_{f(X)}$ and $h_2|_{f(X)}$ to all of $Y$, for they would have to be identically $h_1$ and $h_2$ respectively, which are equal!

What should be done?

I see no value in extending $h_1|_{f(X)}$ and $h_2|_{f(X)}$ to all of $Y$, for they would have to be identically $h_1$ and $h_2$ respectively, which are equal!

The point is that if $f(X)\neq Y$, then it is possible to extend $h_1|_{f(X)}$ to a function that is not equal to $h_1$. That is, Tietze's theorem leads to a proof by contraposition. In particular, there would be a nonzero element of $C(Y)$ that vanishes on $f(X)$.

• Tietze guarantees the existence of $\textit{an}$ extension. Please justify the statement that $h_1|_{f(X)}$ can be extended something not equal to $h_1$. – georgC Dec 9 '11 at 5:03
• @georgC: There exists $y\not\in f(X)$, and you can map $y$ wherever you want, then extend from there. E.g., for the vanishing on $f(X)$ case, there exists $h\in C(Y)$ such that $h(y)=1$ and $h|_{f(X)}=0$. (Edit: I see that Sam's answer made this explicit as well.) – Jonas Meyer Dec 9 '11 at 5:08

If $f$ is not surjective, then - choosing a point $x_0\notin f(X)$ - there is a continuous function $h$ for which $h(x_0) = 1$ and $h|_{f(X)} \equiv 0$.

This follows from Tietze extension and the fact that $\{x_0\}$ and $f(X)$ are closed subsets of $Y$. (The latter because $f(X)$ is compact and $Y$ is Hausdorff).

For this $h$ we have $Cf(h) = Cf(0)$, although $h\ne 0$. So $Cf$ is not injective.

• Ah. Finally, thank you very much. However, what you have invoked here is the Urysohn lemma, not the Tietze extension theorem. I am aware that these two propositions are equivalent, but, again, what you gave was not a straight-forward application of Tietze's theorem. – georgC Dec 9 '11 at 5:13
• @georgC: Please note that $\{x_0\}\cup f(X)$ is closed, and the function that is $1$ at $x_0$ and vanishes on $f(X)$ is continuous on $\{x_0\}\cup f(X)$, then extend it. – Jonas Meyer Dec 9 '11 at 5:16
• @Jonas: Yes, you are correct. When you put it like this, it is most definitely an application of Tietze. Thank you. – georgC Dec 9 '11 at 5:20