= Strict-range modification of a continuous extension
{title2=$\widetilde f=(1-u)f$}
Suppose $Y$ is closed in a <normal space> $X$ and $|f|<1$ on $Y$ for a <continuous> real-valued $f$ on $X$. The <closed set> $Z=\{|f|\ge1\}$ misses $Y$. <Urysohn's lemma> supplies $u:X\to[0,1]$ equal to zero on $Y$ and one on $Z$. Then $(1-u)f$ agrees with $f$ on $Y$ and has range in $(-1,1)$. This permits the inverse transform $s\mapsto s/(1-|s|)$ in the proof of unbounded Tietze extension.
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