I've read the following in a book:
On the projective plane we have $b_1=0$ which suggests that every cycle of edges may be the boundary of a chain of faces. This is, however, not true; taking the sphere with antipodal points identified as our model, consider the path which runs halfway around the equator. This is a cycle and lies in the kernel of $\partial_1$, but is not in the image of $\partial_2$.
My understanding is that the first Betti number $b_1$ is defined as the dimension of $H_1=\ker \partial_1/\text{im } \partial_2$. Here $$\partial_k\colon C_k(T)\to C_{k+1}(T)$$ for a chosen triangulation $T$ of $RP^2$. So if $b_1=0$, then $H_1=\{0\}$ and hence $\ker \partial_1=\text{im } \partial_2$; that is "every cycle of edges may be the boundary of a chain of faces." So I'm not sure what is it that I'm missing. So $b_1=0$ does not imply $H_1=\{0\}$?! Any clarification is appreciated.
Based on a comment to the question: Beware: $\mathbb{R}$-homology (and $\mathbb{Q}$-homology) make torsion vanish. However, $\mathbb{Z}$-homology keeps the torsion mentioned by Lord Shark the Unknown in his comment. So in $\mathbb{R}$-homology, $b_i = 0$ does imply $H_1(\cdot ; \mathbb{R}) \cong \{0\}$.
Here's where the difference comes in: take a chain of edges that "goes around the boundary of a chain of faces twice". (Note that this is slightly fanciful language -- these chains are just $\mathbb{Z}$-weighted sums of edges.) Call this edge chain $c$ and the face chain $b$. So we have $$ 2 \partial b = c \text{.} $$ In integral homology, we're stuck, there is no integer we can mulitply through to clear the coefficient of $\partial b$. In $\mathbb{Q}$ and $\mathbb{R}$, we can multiply through by $\frac{1}{2}$ and discover that $$ \partial b = \frac{1}{2}c \text{.} $$
This literally happens in $\mathbb{R}P^2$ -- there is a $2$-chain whose boundary is not homologous to zero, but two copies of it is. (In homotopy, there is a loop that is not nulhomotopic, but for which two copies is nulhomotopic. This is where I got my fanciful language above.) It turns out this chain generates $H_1$, so $H_1$ is surjected by $k \partial b$ as $k$ ranges over the integers. But every even $k$ is nulhomologous, so $H_1$ only contains $0$ and $\partial b$ and their table of sums is the same as that of $\mathbb{Z}/2\mathbb{Z}$. So we find $H_1(\mathbb{R}P^2;\mathbb{Z}) \cong \mathbb{Z}/2\mathbb{Z}$.
However, in $\mathbb{Q}$ or $\mathbb{R}$ coefficients, we have $\frac{1}{2}$ to simplify the relation and so $\frac{1}{2} \cdot 2 \partial b = 0$ also. This kills all of $H_1$, so $H_1(\mathbb{R}P^2;\mathbb{Q}) \cong H_1(\mathbb{R}P^2;\mathbb{R}) \cong \{0\}$.