Covering map between orientable compact connected surfaces without boundary with genus $g$ Similar to a previous question I asked, this is the other question that I couldn't answer in an exam I gave for my Topology course:
Suppose $g_1,g_2$ are non-negative integers such that $2-2g_2$ divideds $2-2g_1$. I have to construct a covering map $S_{g_1} \rightarrow S_{g_2}$, where $S_g$ is denotes an orientable compact connected surface without boundary with genus $g$.
Now, I attempted to try this problem for surfaces with smaller genera, for example a 2-torus and a 2-sphere. Clearly $2-2g$ is the Euler Characteristic of the said surface, and by classification of surfaces we have to only consider connected sum of 2-tori. So, I have $\chi(S^2)=2$, and $\chi(T^2) = 0$, here we have $2$ dividing $0$, hence I have to construct a covering map from $T^2 \rightarrow S^2$, and I have no idea how to move forward.
Addendum:
I have to got great answers to my question here but I had a doubt when it comes to the points being mapped, specifically in Giulio Bresciani's answer. Clearly the handles of the figure is mapped to the handles in the figure above, but I'm unable to understand the mapping of the twist of the cover (above figure) to the extreme handles of the base (the figure below). Any help?
 A: One image solution: here $g_1-1=2$, $g_2-1=6$, $n=3$. If $g_1$ is greater, just add holes in the middle. If $n$ is greater, just go on with the snake. As pointed out by John Hughes, you clearly need $g_1$ (and hence $g_2$) to be strictly positive. 
 
A: This is a follow up to my comment on John Hughes' answer. The picture below shows how any surface of genus $g \geq 2$ covers the surface of genus $2$. I had to flatten it to draw it so hopefull you understand what I mean in the diagram. If not, looking at the Hatcher reference below will clarify it.

For the more general covering, i.e. $\Sigma_{mn+1} \to \Sigma_m$, you can look at Example $1.41$ on page $73$ of Hatcher. It's the same idea but you 'group' the holes together and then you have an action by $\mathbb{Z}_n$ where $n$ is the degree of the covering. Hope this helps :)
A: Since the sphere has trivial $\pi_1$, the only (connected) covering space of a sphere is a sphere (which is its universal cover!). 
It appears that the statement you were asked to prove is actually false. 
(The torus can be written as a branched cover of the sphere, but that's not what's asked.) 
