# Surjection in compact-open topology [closed]

Let $$Z$$, $$X$$ and $$Y$$ be topological spaces and let $$f:X\to Y$$ be a continuous surjection then is the induced map $$g \to f\circ g$$ from $$C(Z,X)$$ to $$C(Z,Y)$$ is continuous. But is it still a surjection?

My issue is that it's not clear if it has a right-inverse...

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.
• So the sufficient condition is for $f$ to have a continuous right-inverse? – MrMMS May 22 at 16:01
• If $f$ has a right inverse $r$ then $g=f\circ(r\circ g)$... – Jochen Wengenroth May 22 at 16:37
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 $$\pi_1$$.