Strict-range modification of a continuous extension ($\tilde{f}=(1-u)f$)

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 $Z=\{|f|\ge 1\}$ 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.