Let $A$ be a ring, $B$ a finite projective $A$-algebra, and $C$ a finite projective $B$-algebra. We can show that $C$ is also finite and projective when regarded as an $A$-algebra (by, for instance, taking direct sum with appropriate $A$-module to make it free in $(A\text{-Mod})$.)

The following are what I would like to prove / figure out:

(1) The map $$\hom_A(B,A)\otimes_B \hom_B(C,B)\to\hom_A(C,A),\;\;\; f\otimes g\mapsto f\circ g$$ is a surjection. Note that the $\hom_A$ stands for $A$-linear maps (instead of $A$-algebra maps). So the statement makes sense if we only require $C$ to be a $B$-module (instead of a $B$-algebra).

(2) Express $[C:A]$ in terms of $[C:B]$ and $[B:A]$. Here $[C:A]$ is the map $$[C:A]:\text{Spec }A\to\mathbb Z,\;\;\;\mathfrak p\mapsto\text{rank}_{A_\mathfrak p}C_\mathfrak p.$$ ($[C:B]$ and $[B:A]$ are defined in the same way.) Note that $B_\mathfrak p$ is free over $A_\mathfrak p$, the rank is just the "free rank".

For (1), I localize the map at a prime ideal $\mathfrak p$ to see whether it is surjective. Since both $B_\mathfrak p$ and $C_\mathfrak p$ are free over $A_\mathfrak p$, we can write $C_\mathfrak p \simeq A_\mathfrak p ^n$ and $B_\mathfrak p\simeq A_\mathfrak p ^m$ as $A_\mathfrak p$-modules. On the other hand, since $B_\mathfrak p=B\otimes_A A_\mathfrak p$ and $C_\mathfrak p=C\otimes_A A_\mathfrak p$, we know $C_\mathfrak p$ is projective (but not necessarily free) $B_\mathfrak p$-module. Suppose in $(B_\mathfrak p\text{-Mod})$that $C_\mathfrak p \oplus D \simeq B_\mathfrak p^k$. Then in $(A_\mathfrak p\text{-Mod})$ we have $C_\mathfrak p\oplus D\simeq A_\mathfrak p ^{mk}$. Now, given any $A_\mathfrak p$-linear map $C_\mathfrak p\to A_\mathfrak p$, I tried to decompose it into a sequence of maps $$C_\mathfrak p\to C_\mathfrak p \oplus D \simeq B_\mathfrak p^k \to B_\mathfrak p \simeq A_\mathfrak p^ m\to A_\mathfrak p$$ so that $C_\mathfrak p\to B_\mathfrak p^k\to B_\mathfrak p$ is $B_\mathfrak p$-linear. But the isomorphism $C_\mathfrak p \oplus D \simeq B_\mathfrak p^k$ is so unclear that I do not know how to write down the maps in the sequence.

For (2), I have even less idea to get started. I have just drawn a diagram about the maps, and to guess from the free case that we should have something like $$\text{(??) }\text{rank}_{A_\mathfrak p}C_\mathfrak p = \text{rank}_{B_\mathfrak p} C_\mathfrak p\cdot \text{rank}_{A_\mathfrak p} B_\mathfrak p$$ or $$\text{(??) }[C:A](\phi^{-1}(\mathfrak q))=[B:A](\phi^{-1}(\mathfrak q))\cdot [C:B](\mathfrak q), $$ where $\mathfrak q$ is a prime ideal in $B$ and $\phi: A\to B$ is the natural map which determent the $A$-algebra structure on $B$. But both of the two identities seem to be problematic.