In Griffiths-Harris, Principles of Algebraic Geometry, page 457, it says that the kernel of the first Chern class $c_1:Pic(M)\cong H^1(M,\mathcal O^*)\longrightarrow H^2(M,\mathbb Z)$ is the space of divisors homologous to $0$ (everything up to linear equivalence).
Why is this true? The kernel of $c_1$ should be $Pic^0\subset Pic$, but I fail to see why it corresponds to homological equivalence, and therefore why $$rk ({\rm Im} \ c_i)=rk\frac{\mbox{divisors on $M$}}{\mbox{homological equivalence}}.$$ The answer to the question Neron-Severi group as the image of first Chern class does not solve the problem, unfortunately.
Thank you in advance.