Surjection in compact-open topology

Let $X = Y = S^1 = \mathbb{R}/ \mathbb{Z}$ be the circle and let $f \colon X \to Y$ be given by $f(t) = 2t$ which is continuous and surjective.

There exists no $g \in C(S^1,X)$ such that $f \circ g = \mbox{Id}_{S^1}$, as $f$ induces the doubling map on $\pi_1$.

In many cases, $f_\ast: C(Z,X)\to C(Z,Y)$, $g\mapsto f\circ g$ is not surjective: Put $Z=Y$ and $h=id_Y \in C(Z,Y)$, then $h$ is in the range of $f_\ast$ only if $f$ has a right inverse.