Is the map $z\mapsto z^2$ a fibration?

Question

Let $f:\mathbb C \to \mathbb C$ be defined as $f(z) = z^2$.

Is this a fibration? Or at least a Serre fibration?

I am not sure how to approach this.

Answer 1

Connor's comment sums up the situation.

Two fibres of a Hurewicz fibration over points in the same path-component of the base are homotopy equivalent.

Thus $f:\mathbb{C}\rightarrow\mathbb{C}$, $z\mapsto z^2$, is not a Hurewicz fibration, since the origin has a unique preimage, whilst all other points have two.

A similar statement holds for Serre fibrations when homotopy equivalent is replaced by weakly homotopy equivalent. Thus $f$ is not even a Serre fibration.

On the other hand, by restriction we obtain $g:\mathbb{C}\setminus\{0\}\rightarrow\mathbb{C}\setminus\{0\}$, $z\mapsto z^2$, and this map is locally-trivial. Since $\mathbb{C}\setminus\{0\}$ is paracompact $T_2$, this map is thus a Hurewicz fibration. The fact that $g$ is locally trivial is easy to see. The easiest way to do so is to use polar coordinates to identify $\mathbb{C}\setminus\{0\}\cong S^1\times(0,\infty)$ and notice that $g$ is the product of the two-fold covering projection $S^1\rightarrow S^1$, $z\mapsto z^2$, and the homeomorphism $(0,1)\rightarrow (0,1)$, $t\mapsto t^2$.