The onset temperature ${T}_{0}(P)$ for superfluidity of $^{4}\mathrm{He}$ has been measured in $d=0.1\ensuremath{-}, 0.2\ensuremath{-}, 0.4\ensuremath{-}, \mathrm{and} 0.6\ensuremath{-}\ensuremath{\mu}m$ wide channels at pressures between saturated vapor pressure and the melting curve. The onset was detected by observing the vanishing of second sound in resonators equipped with superleak transducers. In the channels used the onset temperature ${T}_{0}(P)$ was depressed below the bulk transition temperature ${T}_{\ensuremath{\lambda}}(P)$ by $2\ifmmode\times\else\texttimes\fi{}{10}^{\ensuremath{-}5}\ensuremath{\le}{\ensuremath{\epsilon}}_{0}\ensuremath{\le}2\ifmmode\times\else\texttimes\fi{}{10}^{\ensuremath{-}4}$. Here ${\ensuremath{\epsilon}}_{0}=1\ensuremath{-}\frac{{T}_{0}(P)}{{T}_{\ensuremath{\lambda}}(P)}$. For constant $d$, we find that ${\ensuremath{\epsilon}}_{0}$ is independent of pressure to within our resolution of \ifmmode\pm\else\textpm\fi{}2%. From the data the exponent ${\ensuremath{\lambda}}^{\ensuremath{'}}$ and the coefficient ${d}^{*}$ for the equation ${\ensuremath{\epsilon}}_{0}={(\frac{d}{{d}^{*}})}^{\ensuremath{-}{\ensuremath{\lambda}}^{\ensuremath{'}}}$, and the exponent and coefficient for the correlation length, $\ensuremath{\xi}={\ensuremath{\xi}}^{*}{\ensuremath{\epsilon}}^{\ensuremath{-}{\ensuremath{\nu}}^{\ensuremath{'}}}$, have been calculated. We show that ${\ensuremath{\lambda}}^{\ensuremath{'}}$ and ${d}^{*}$, as well as ${\ensuremath{\nu}}^{\ensuremath{'}}$ and ${\ensuremath{\xi}}^{*}$, and the quantity $\frac{\ensuremath{\xi}{\ensuremath{\rho}}_{s}}{T}=(0.16\ifmmode\pm\else\textpm\fi{}0.01)\ifmmode\times\else\texttimes\fi{}{10}^{\ensuremath{-}8}$ are independent of pressure, within our resolution of \ifmmode\pm\else\textpm\fi{}2%. The lack of a pressure dependence of the critical exponents is in agreement with universality arguments. For the absolute values we find $\ensuremath{\xi}=(1.2\ifmmode\pm\else\textpm\fi{}0.1)\ifmmode\times\else\texttimes\fi{}{10}^{\ensuremath{-}8}{\ensuremath{\epsilon}}^{\ensuremath{-}0.65\ifmmode\pm\else\textpm\fi{}0.02}$, ${\ensuremath{\epsilon}}_{0}={[\frac{d}{(5.7\ifmmode\pm\else\textpm\fi{}0.6)\ifmmode\times\else\texttimes\fi{}{10}^{\ensuremath{-}8}}]}^{\ensuremath{-}1.54\ifmmode\pm\else\textpm\fi{}0.05}$, and $\frac{\ensuremath{\xi}{\ensuremath{\rho}}_{s}}{T}=(0.16\ifmmode\pm\else\textpm\fi{}0.01)\ifmmode\times\else\texttimes\fi{}{10}^{\ensuremath{-}8}$.