Standard Big Bang model and space curvature
What exactly did the book say? It all depends on what you mean/define as "curvature". What you describe appears to be a description of the behaviour of $\Omega$. Inflation does indeed drive $\Omega$ towards unity and simultaneously flattens space because the radius of curvature grows exponentially bigger. 

If $\Omega &lt; 1$ at some early epoch then it should decrease quickly with time such that $\Omega &lt;&lt; 1$ in the present day - this means that the universe has negative curvature, but does not mean it is becoming more curved.

In the Friedmann equation, the curvature parameter $k$ is a constant $(1,0,-1)$
$$H^2 = \frac{8\pi G\rho}{3} - \frac{kc^2}{a^2}$$

Here, the spatial curvature is $k/a^2$ and the radius of curvature is $a$ if $k=+1$. Thus as the universe expands and $a$ gets larger, any curvature becomes smaller.

In a little more detail - one can write the above equation in terms of the density parameter $\Omega$, the ratio of density to the critical density $3H^2/8\pi G$:
$$(\Omega ^{-1}-1)\rho a^{2}={\frac {-3kc^{2}}{8\pi G}}$$

During inflation, the energy density $\rho c^2$ remains constant as $a$ grows exponentially. In order to keep the left hand side equal to the right hand side (which is just a collection of constants), then $\Omega$ must be driven very close to unity, while $k/a^2$ will tend towards zero.

After inflation then $\rho$ will vary with $a$ depending on whether the expansion is dominated by matter ($a^{-3}$) or radiation $(a^{-4})$. In both these cases $\rho a^2$ will decrease as the universe expands, such that if $k \ne 0$, then $(\Omega^{-1} -1)$ must increase, which means that $\Omega$ must either grow or shrink away from unity. But $k/a^2$ continues to get smaller as the universe expands.