Non-Homeomorphicity of Compact surfaces Using the fact that every compact surface is homeomorphic to either a connected sum of torii or a connected sum of real projective planes, and using the fact that the corresponding fundamental groups are 
$$\pi_1(\Sigma_g)=\langle a_1,...,a_g,b_1,...,b_g|[a_1,b_1]\cdot...\cdot[a_g,b_g]=1\rangle$$
$$\pi_1(S_g)=\langle c_0,...,c_g|c_0^2\cdot...\cdot c_g^2=1\rangle$$
And I want to show that these fundamental groups are always non isomorphic. My idea was to surject them onto some groups using the Universal property of presentations and get some contradiction. In practice though, this seems pretty hard. I can see an easy surjection from $\pi_1(\Sigma_g)\to\mathbb{Z}^{2g}$, but nothing else.
 A: The key here is to abelianize (which essentially means look at $H_1$ and is exactly the surjection onto $\mathbb{Z}^{2g}$ you were talking about). Note that the product of commutators for the orientable surface when abelianized yields a trivial relation because $$a_1b_1a_1^{-1}b_1^{-1} \mapsto a_1 + b_1 - a_1 - b_1 = 0$$ So the abelianization of $\pi_1(\Sigma_g) = H_1(\Sigma_g) = \mathbb{Z}^{2g}$. So if $\Sigma_g \not \cong \Sigma_{g'}$ then $\pi_1(\Sigma_g) \not\cong \pi_1(\Sigma_{g'})$. 
Now let us consider what happens when we abelianize $\pi_1(S_g)$. The relation $c_0^2...c_g^2 = 1$ turns into $2c_0 +...+ 2c_g = 0$, which I can write as $2(c_0 + ... + c_g) = 0$. So there is a non-zero element with $2$-torsion in this abelianization, which is not true in the abelianization of $\pi_1(\Sigma_g)$. So $\pi_1(\Sigma_g) \not \cong \pi_1(S_{g'})$. 
Note that we can rewrite the presentation for the abelianized group now by replacing $c_g$ with $c_0 + ... + c_g$. Basically, instead of saying $\pi_1(S_g)$ is the free abelian group with generators being $c_0,...,c_g$ and relator being $2(c_0+..+c_g) = 0$, we can instead write it as the free abelian group on $d_1,...,d_g$ with relator $2d_g = 0$, where $d_i = c_i$ for $i < g$ and $d_g = c_1+...+c_{g-1}$. This shows us that the abelianization of $\pi_1(S_g) = H_1(S_g) = \mathbb{Z}^{g-1} \oplus \mathbb{Z}_2$. Thus if $S_g \not \cong S_{g'}$ then $\pi_1(S_g) \not \cong \pi_1(S_{g'})$. 
Edit: I wasn't sure if you're familiar with homology so if not, ignore any mention of $H_1$. Also, if you're not familiar with abelianization, I can rewrite this to be more explicit!
