Show that if \(A \hookrightarrow X\) is a cofibration of compact Hausdorff spaces, then for any space \(Y\), the map \(Y^X \to Y^A\) obtained by restriction of functions is a fibration.
Solution
Let \[ i:A\hookrightarrow X \] be a cofibration, with \(A\) and \(X\) compact Hausdorff, and let \[ r:Y^X\longrightarrow Y^A \] be restriction. We use the compact-open topology on mapping spaces.
To verify the homotopy lifting property for \(r\), suppose given a space \(Z\), a map \[ f_0:Z\to Y^X, \] and a homotopy \[ g:Z\times I\to Y^A \] with \[ g(z,0)=r(f_0(z)). \] By the exponential law for compact Hausdorff source spaces, these data are adjoint to continuous maps \[ F_0:Z\times X\to Y, \qquad G:Z\times I\times A\to Y \] which agree on \(Z\times\{0\}\times A\).
Since \(A\hookrightarrow X\) is a cofibration, so is \[ Z\times A\hookrightarrow Z\times X. \] Equivalently, the inclusion \[ (Z\times X\times\{0\})\cup(Z\times A\times I) \hookrightarrow Z\times X\times I \] has the extension property for maps to arbitrary spaces. Define a map on this subspace by \[ (z,x,0)\longmapsto F_0(z,x), \qquad (z,a,t)\longmapsto G(z,t,a). \] The two formulas agree on the intersection, so they extend to a continuous map \[ \widetilde G:Z\times X\times I\to Y. \] Adjointing again gives a homotopy \[ \widetilde g:Z\times I\to Y^X \] with \(\widetilde g(-,0)=f_0\) and \[ r\widetilde g=g. \] Thus \(r\) has the homotopy lifting property for every \(Z\), hence \[ \boxed{Y^X\to Y^A\text{ is a fibration}.} \]