4
$\begingroup$

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...

$\endgroup$
5
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ So the sufficient condition is for $f$ to have a continuous right-inverse? $\endgroup$ – MrMMS May 22 at 16:01
  • $\begingroup$ If $f$ has a right inverse $r$ then $g=f\circ(r\circ g)$... $\endgroup$ – Jochen Wengenroth May 22 at 16:37
  • 1
    $\begingroup$ Yes, of course, Flabby. $\endgroup$ – Todd Trimble May 22 at 17:32
5
$\begingroup$

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 $\pi_1$.

| cite | improve this answer | |
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.