What's the generator of the n-th homology of $RP^n$ for $n$ odd? I know that $H_n(\mathbb{R}P^n)=\mathbb{Z}$ for $n$ odd, I am just wondering the generator of it, I thought the canonical map $\pi: S^n\to \mathbb{R}P^n$ might be helpful.
 A: I'm not sure if this is satisfying, but viewing $\mathbb RP^n$ as a CW complex, we take the attaching map of the disk $D_k$ with attaching map $\phi_k:S^{k-1} \to \mathbb R P^{k-1}$, where this is the covering map, for all $k \leq n$. But then we can view $S^{k-2} \hookrightarrow S^{k-1}$ as the equator, and note that  we can check the composition $ \rho \circ \phi$ where $\rho:\mathbb RP^{k-1}/\mathbb RP^{k-2} \cong S^{k-1}$, and check that what is going on is that the preimage of $\mathbb RP^{k-2}$ is the equator, so two open balls are being mapped to the quotient, one ball is mapped antipodally, and the other by identity (they differ by an antipodal action.)
The basic point here is that the map has degree $0$ if $k$ is odd and $2$ if $k$ is even.
To me, this tells that we can take the quotient of the chain  complex 
$$0 \to \mathbb Z \to \dots \mathbb Z \to 0$$
alternating by zero maps and $\times 2$ maps.
Hence, we have that $\mathbb RP^n= \mathbb RP^{n-1} \cup e_n$ and $\partial_n(e_n)=2e^{k-1}$ when $n$ is even, and the zero map with $n$ odd. so the kernel is all of $\mathbb Z$, while there is no $e_{n+1}$, so the generator for homology is basically just $[e_n]$ when taking the $CW$ point of view.
A: This is an ellaboration of Tyrone's comment.
Let $M$ be a compact oriented smooth manifold of dimension $n$ (here by orientation I mean a compatible orientation of each tangent space). Take a triangulation of $M$ and order the vertices of each $n$-simplex according to the orientation of $M$ , that is: for each $n$-simplex $\sigma$


*

*take the inverse image of $\sigma$ under the exponential map at some vertex $x$ of $\sigma$.

*order the vertices of $\sigma$ so that $x=x_0$ and $(x_0x_1,x_0x_2,\ldots,x_0x_n)$ is an oriented basis of $T_{x}M$


Now one proves that the boundary of the sum of these ordered simplices (each with coefficient $1$) is zero (this requires a computation).
This is a representative of the fundamental class of your manifold, which is a generator of the top homology group.
In particular, for $RP^3$, you may take the octahedral triangulation of the sphere $S^3$, and then remove one of the vertices and its open star. You are left with $2^3$ simplices, whose image under the quotient map triangulates $RP^3$.
