$$H^1(X) = H^1(Y)^{<\pm 1>} =
H^1(E)^{<\pm 1> } = 0, \qquad\text{and}$$ $$H^{2,0}(X) = H^{2,0}(Y \times E)^{<\pm 1>} =
H^2(Y)^{<\pm 1>} = 0.$$ Thus $X$ is indeed a smooth Calabi–Yau threefold. It is non-rigid, because $T(Y\times
E)$ descends to $X$ and each of its Hodge pieces $H^{p,q}(T(Y\times E))$ is $1$-dimensional.

To compute the monodromy around each singular fiber, we notice that for a generic fiber $X_t =
f^{-1}(t)$ the Kummer structure gives a canonical decomposition $$H^2(X_t,{{\mathbb Q}}) =
(H^2(A_t,{{\mathbb Q}}) \oplus {{\mathbb Q}}A_t[2])^{<\pm 1>}.$$ Our examples were chosen so that the action of $\pm 1$ on $A_t[2]= E_{\Gamma,t}[2] \times
E_t[2]$ is trivial. Moreover, in the Künneth decomposition $$H^2(A_t) = H^2(E) \oplus (H^1(E)\otimes
 H^1(E_{\Gamma,t})) \oplus H^2(E_{\Gamma,t})$$ the $\pm 1$ action is trivial on the first and last factors, is trivial on $H^1(E)$ and is unipotent on $H^1(E_\Gamma,t)$ around each singular fiber of $f$ (namely the cusps of $\Gamma$). Thus the monodromy of the fibration $f$ is unipotent as well.

To compute the Kodaira–Spencer map $\Theta(f)$ for our $f$ we embed it into the Kodaira–Spencer map for $Y\times E \rightarrow
X(\tilde{\Gamma})$. This map is the cup product with the Kodaira–Spencer class $\Theta$ which itself is $\Theta_{Y/X(\tilde{\Gamma})} \otimes
\Theta_{X(\tilde{\Gamma}) \times
E/X(\tilde{\Gamma})}$. Since the Kodaira–Spencer 