Let us work over $\mathbb C$. In this article by S. Katz, it is stated that for a quintic threefold $X\subset \mathbb P^4$ one has $$H_2(X,\mathbb Z)\cong\mathbb Z.$$ Can anyone help me to see why this is true?
Remark. If $X$ was a general surface, this would be easier, as we would have $$H_2(X,\mathbb Z)\cong H^2(X,\mathbb Z)=\textrm{Pic}(X)=\mathbb Z.$$ The last equality uses generality (I guess this is really necessary, but I lack a counterexample). That is why I would also accept an answer assuming the threefold $X$ to be general. However, in the quoted paper the result is stated with no genericity assumption, so it should be true as stated.
The group in question, $H_2(X,\mathbb Z)$, is now isomorphic to $H^4(X,\mathbb Z)$. I do not find it easier to work with the latter, though.
I do not know if I can use the exponential sequence, as taking its cohomology would give me a sequence of vector spaces, and the group $H^4(X,\mathbb Z)$ does not even appear there. Any hint or solution would be much appreciated.
I was also wondering whether $H_2(X,\mathbb Z)\cong\mathbb Z$ can hold for other projective Calabi-Yau threefolds, different from the quintic in $\mathbb P^4$.
Thank you!
One simple and powerful answer: Lefschetz Hyperplane Theorem.