The difference between Urysohn's lemma and Tietze Extension Theorem.

Urysohn's lemma says that if $X$ is a normal space, then for every two disjoint closed sets $F_{1},F_{2}\in X$, there exists a continuous function $f:X\to [a,b]\in\Bbb{R}$ such that $f(F_{1})=\{a\}$ and $f(F_{2})=\{b\}$.

Tietze Extension theorem says that for every such $f$, there exists a continuous function $f^*:X\to [a,b]$ such that $f^*|F_{1}$ and $f^*| F_{2}=f$.

I don't understand the difference! Why can't $f^*=f$? And if we assume $f^*\neq f$, are we just saying that there are two such continuous functions of the type mentioned in Urysohn's lemma?

• I don't know where you're getting these statements from, but "every such $f$" is just wrong. – Chris Eagle Jun 26 '13 at 16:02
• "every such $f$" as in every such $f$ as constructed in Urysohn's lemma. This has been quoted verbatim. – fierydemon Jun 26 '13 at 16:05
• Then your book is completely wrong; throw it away. – Chris Eagle Jun 26 '13 at 16:06

One can indeed obtain the result of UL via TET by setting $f|_{F_1} = a$ and $f|_{F_2} = b$. Also, in this Wikipedia article it is explicitly written that TET generalizes UL.
• @AyushKhaitan: Urysohn's lemma uses two non-empty disjoint closed sets. What you wrote in your question is just an application of Tietze's extension theorem to two closed sets. But the theorem itself says that each continuous $f$ from a closed set $C$ to $\mathbb R$ can be extended to a continuous $\hat f$ from the entire space $X$ to $\mathbb R$. – Stefan Hamcke Jun 26 '13 at 16:03