Does homology depend on the topology?

Let $S_g$ be a closed oriented surface of genus $g\ge 2$. I know that its singular homology groups are $H_0(S_g,\mathbb{R})=H_2(S_g,\mathbb{R})=\mathbb{R}$ and $H_1(S_g,\mathbb{R})=\mathbb{R}^{2g}$.

My question is: do they depend on the topology on $S_g$?

On $S_g$ we could consider different riemannian metrics (hyperbolic, or flat with conical singularities for example) which induce different topologies. Do all of them produce the same singular homology groups? Why?

I guess the answer should be "yes", but this feels strange to me, since the definition of singular $n-$simplex concerns $continuous$ maps from $\Delta^n$ to the space we are considering

• Different Riemannian metrics on $S_g$ all produce the same topology. – Cheerful Parsnip Jun 13 '17 at 13:16
• @GrumpyParsnip Thank you for your comment, I didn't know that. Why is that? Where can I find a proof of this fact? – user384588 Jun 13 '17 at 13:17
• It should follow from the classification of surfaces up to homeomorphism. There's only one closed orientable surface for each genus $g$. – Cheerful Parsnip Jun 13 '17 at 13:22
• I'm sort of confused, what kind of a thing is $S_g$ if not a topological space? – s.harp Jun 13 '17 at 13:36
• @user384588: To elaborate on s.harp's comment, normally we think of a manifold as first and foremost being a topological space, and then we put additional structure like a differentiable structure and a Riemannian metric. The topology doesn't come from the Riemannian metric; it exists prior to defining a Riemannian metric at all. – Eric Wofsey Jun 13 '17 at 21:45

To expand on the comment of @GrumpyParsnip, one of the basic theorems of Riemannian geometry is that for any Riemannian manifold $M$ and any $p \in M$ there exists $r>0$ such that the exponential function $e_p : T_p M \to M$ restricts to a homeomorphism from $U_p(r) = \{$the $r$-neighborhood of the origin in $T_p M$ with respect to the given Riemannian metric$\}$ to an open subset of $M$ containing $p$. It follows that the collection of open subsets $\{e_p(U_p(s)) \,\bigm|\, 0<s<r\}$ forms a neighborhood basis of $p$ in $M$. Since this is true independent of the Riemannian metric, it follows that the topology induced by the Riemannian metric equals the given topology on $M$.