Path-connectedness of $\mathbb{C}\setminus \lbrace z_1, ..., z_n \rbrace$ Today I was thinking about the fact that $\mathbb{C} \setminus \lbrace z_1, ..., z_n \rbrace$ is path-connected. I know how to prove that, just constructing a path joining two points, but I was wondering : is there a way to prove this using the fact that the image of a path-connected space by a continuous function is path-connected ?
More precisely, it is possible prove that $\mathbb{C} \setminus \lbrace z_1, ..., z_n \rbrace$ is path-connected by defining explicitely a surjective continuous function $f : X \rightarrow  \mathbb{C} \setminus \lbrace z_1, ..., z_n \rbrace$, where $X$ is a path-connected topological space ?
For the case $n=1$, using $\exp : \mathbb{C} \rightarrow \mathbb{C}^*$ works, but I was not able to generalize that in the case you remove more than one point.
Thanks for your help !
 A: First of all, the explanation "just constructing a path joining two points" proves absolutely nothing. I suggest, you write a careful proof of path-connectedness of $Z^c$ which is the complement to a given finite subset $Z\subset {\mathbb C}$ and compare complexity of the proof to the complexity of your proof that ${\mathbb C}^*$ is path-connected (it hinges upon proving continuity and surjectivity of the complex exponential function $\exp: {\mathbb C}\to  {\mathbb C}^*$: Proving this from scratch is a nontrivial challenge).
No, let's prove path-connectivity of $Z^c= {\mathbb C}\setminus Z$ along the lines of your request.

*

*First, consider a finite subset $W\subset  {\mathbb R} \subset {\mathbb C}$. Then $W^c$ is the union of two subsets $U^+, U^-$, where
$$
U^+=\{z: Im(z)\ge 0\}\setminus W$$
and
$$
U^-=\{z: Im(z)\le 0\}\setminus W.$$
In fact, both sets are homeomorphic via the complex conjugation map. I claim that $U^+$ is path-connected. Indeed, it is star-like with respect to the point $i$: For each $z\in U^+$ the line segment $[i, z]$ between $i$ and $z$ is contained in $U^+$. I will leave it for you to verify that each star-like subset of  ${\mathbb C}$ is path-connected. (Concatenate two segments $[p, i], [i,q]$ to connect points $p, q\in U^+$.) It follows that $U^-$ is path-connected as well. Observe that $U^+\cap U^-\ne \emptyset$, which implies path-connectivity of  the union $U^+\cup U^-=W^c$.


*Second, I will use the fact that if $W, Z$ are two subsets of the same cardinality in ${\mathbb C}$ then there exists a homeomorphism $h: {\mathbb C}\to {\mathbb C}$ sending $W$ bijectively to $Z$, see for instance here for an explicit construction.
Thus, given aby finite subset $Z\subset {\mathbb C}$, I find a finite subset $W\subset 
 {\mathbb R} \subset {\mathbb C}$ of the same cardinality and, hence, a homeomorphism
$$
f: X:= W^c\to Z^c
$$
obtained by restricting the homeomorphism $h$. This is the continuous surjective function from a connected space $X$ to $Z^c$ you asked for.
Needless to say, this is much more complex than the standard direct proof of path-connectivity of $Z^c$.
