# Covering a connected sum

I have the following problem as a part of my homework:

Let $S$ be a closed surface (compact and connected). Show that for every $k$ exists a covering map of $k$ folds $p_k:S_k \rightarrow \mathbb{T}\sharp S$, where $\mathbb{T}$ is the torus and $S_k$ is the closed surface with the same orientability as $S$ and $\chi(S_k) = k \cdot \chi (\mathbb{T}\sharp S)$.

I've done this: as a first approach, suppose $S$ is the sphere $\mathbb{S}^2$. Then $S$ is orientable and $\mathbb{T}\sharp S = \mathbb{T} \sharp \mathbb{S}^2 = \mathbb{T}$, so $k \cdot \chi(\mathbb{T} \sharp S) = k \cdot \chi(\mathbb{T}) = k \cdot 0 = 0$, and $S_k$ has the same orientability as $S$ (orientable) and $\chi(S_k) = 0$, so necessarily $S_k = \mathbb{T}$. This means that we must find, for every $k$, a covering map $p_k:\mathbb{T} \rightarrow \mathbb{T}$ with $k$ folds. If we consider $\mathbb{S}^1$ as the subset $\{z \in \mathbb{C} : |z|=1\}$ of $\mathbb{C}$, then $\mathbb{S}^1 \rightarrow \mathbb{S}^1$ sending $z$ in $z^k$ is a $k$-folded covering for $S^1$, and if we write $\mathbb{T} = \mathbb{S}^1 \times \mathbb{S}^1$, we can construct the map $p_k: \mathbb{S}^1 \times \mathbb{S}^1 \rightarrow \mathbb{S}^1 \times \mathbb{S}^1$ with $p_k(z_1,z_2) = (z_1^k,z_2)$ which gives the $k$-folded covering.

Now, if $S$ is whatever, using Euler's characteristic is easy to show that $S_k$ must be the connected sum of a torus and $k$ copies of $S$, this is, $S_k = \mathbb{T} \sharp S \sharp ...^{k)} \sharp S$.

So we must find a $k$-folded covering $p_k:\mathbb{T} \sharp S \sharp ...^{k)} \sharp S \rightarrow \mathbb{T} \sharp S$. It seems the easiest way would be the following: use the $k$-folded covering of the torus we already have in the torus, and stack the $k$ copies of $S$ onto $S$ by an identity-like map, so that given $x \in S$, $p_k^{-1}(x)$ consists in $k$ copies of that point, one in each copy of $S$. But we can't make it that way directly, beacause the group homomorphism between fundamental groups induced by a covering map is injective, so we cant define the coverings for the torus and $S$ separately: if $S$ is, for instance, a torus, a covering map from $k$ glued torus into a torus would induce an injective homomorphism between $\mathbb{Z}^k$ and $\mathbb{Z}$, which is not possible.

I would like a clue in order to build the covering map. I can't use many arguments about fundamental groups or covering spaces clasification theorems. A more general version of this problem is 11-5 in Lee, "Introduction to topological manifolds", but I haven't found anything about it.

Thank you.

• Isn't your map $p_k:T\rightarrow T$ a $k^2$ fold covering? For example, when $k=2$, $p_2^{{-1}}(1,1) = \{(\pm 1, \pm 1)\}$. I'll keep thinking about the rest...... – Jason DeVito Apr 2 '12 at 20:08
• Alright, first, to repair your $p_k$ map, just do the $z^k$ thing in one coordinate and leave the second alone. Second, at least in the orientable case, Hatcher has a great picture and explanation on page 73 of his Algebraic Topology book. The nonorientable case may work similarly. – Jason DeVito Apr 2 '12 at 20:40
• Yes, you're right about the $p_k$ map. I'll have a look at Hatcher's. – rf1x Apr 2 '12 at 21:28
• I think Hatcher's idea will work fine in the nonorientable case as well, roughly the idea is as follows. Start with a torus. Add $k$ equally spaced "spokes", where each spoke is a connect sums with $S$. Then there's a natural free $\mathbb{Z}/k\mathbb{Z}$ action rotating the whole picture. The quotient space of this action, at least pictorally, is $S\sharp T$, as it should be. – Jason DeVito Apr 2 '12 at 22:41
• @JasonDeVito Yes, it works. Here's a pic of what I did. It's in spanish, but I think the draws will be enough. cl.ly/3L2O1R0s281s0J0d2h2G. Thank you very much. – rf1x Apr 2 '12 at 23:14