Linearly Independent Curves on Genus g Surface I have seen the following claim:
Let $\gamma_1, \gamma_2, \cdots, \gamma_g$ be a collection of $g$ non-intersecting, simple closed curves on a genus $g$ surface, $\Sigma$. Then the $\gamma_i$ are linearly independent (as elements of the first homology) if and only if $\Sigma - \gamma_1 \cup \cdots \cup \gamma_n$ is connected.
I feel like this should be easy to see, but I cannot seem to come up with a satisfactory proof (I have tried the standard things like Mayer-Vietoris, LES). I was hoping there was an algebraic topology-type proof (as I have been trying), but maybe not? If someone could help me out, I would be very grateful :).
 A: Let $\Sigma$ be a closed orientable surface of genus $g$. I will assume PL or DIFF as the underlying category, although for dimension $\leq 3$ those coincide with TOP.
Let $\gamma_i$ be the embeddings of $S^1$ and $[\gamma_i]$ denote the image of their fundamental class in $H_1(\Sigma)$.
Suppose $[\gamma_i]$ are linear independent. If $\Sigma - \cup \gamma_i$ was not connected, the homology class represented by a subset of the simple loops (up to signs) would be nullhomologous, hence condradiction. For this claim we need the fact, that the fundamental class of the boundary of a compact orientable manifold $M$ is nullhomologous in $H_{n-1}(M)$. This can be seen by an easy stratifold argument or by an explicit simplicial construction (the latter in low dimensions $\leq 2$ is rather intuitive but also $\leq 3$ should work). A third way to see this would be to consider the long exact sequence of the pair $(M,\partial M)$. 
You will get $H_n (M,\partial M) \stackrel \partial \to H_{n-1}(\partial M) \to  H_{n-1}(M)$ and then use the fact that the boundary operator $\partial$ send fundamental class to fundamental class, where consequently the latter will be in the kernel of the inclusion map $i$.
For a rigorous attempt, why this contradicts the LHS:
Let $\hat \Sigma \subset \Sigma$ be a closed subset for which its interior becomes a component of $\Sigma - \cup \gamma_i$. Then there exists a subset (of indices) $I \subset \{1,\cdots ,g\}$, such that $L = \bigcup_{i\in I} \epsilon_i \gamma_i$ is the boundary of the orientable manifold $\hat \Sigma$. Here  $\epsilon_i \in \{+,-\}$ in the sense of orientation. But then the inclusion $L \to \Sigma$ factors through $(L,\emptyset) \to (\hat \Sigma ,L) \to (\Sigma,\emptyset) $ and by functoriality the induced map on $H_1$ sends the fundamental class of $L$ to zero, hence $\sum_{i \in I} \epsilon_i [\gamma_i]=0$. (the functoriality argument works, because we have shown above that $L \to (\hat \Sigma,L)$ induces a map on homology with this property).
For the other implication we want to consider the subgroup $G \subset \pi_1(\Sigma)$ generated by the $[\gamma_i]$. By covering space theory, there exists a covering space corresponding to this subgroup $G$. Such a covering space (which is unique up to covering isomorphism) can also be obtained by cutting along the embedded spheres $\gamma_i$, taking $\mathbb Z^g$ copies of the connected complement and glue them together under the corresponding bicollared loops. It is important to choose such a bicollar, not only to stay in the category, but also because this means that you get on the complement "boundary" $g$ + signs and $g$ - minus signs. So you can construct the covering space through some kind of a lattice-gluing. Kt is easy to see that $Deck \cong \mathbb Z^g$ and therefore, since rank is preserved in short exact sequences, also $G \cong \mathbb Z^g$. This implies the linear independence of the $[\gamma_i]$.
I somehow rushed over the last part due to a meeting I am late to. If you have further questions, I will be happy to edit it and go more into detail. You could find details on the construction for example in Burde & Zieschang's "Knots" Book in Chapter 4.2 if i remember correctly.
Hope this helps although some details were left open.
