I'm reading Logic as Algebra and the definition for a Boolean subalgebra of a Boolean algebra $A$ is a subset $B$ of $A$ such that $B$, with respect to the operations of $A$, is a Boolean algebra.
This seems fair.
Immediately an example of a Boolean algebra $B$ which is a subset of a Boolean algebra $A$ such that $B$ is NOT a Boolean subalgebra of $A$ is given. Namely, if $X$ is an arbitrary set and $A=\mathcal{P}(X)$, and $Y$ is a proper subset of $X$ then $B=\mathcal{P}(Y)$ is a subset of $A$ and is itself a Boolean algebra but not a Boolean subalgebra of $A$. The operations are the natural ones ($1_A = X$, $0_A=\emptyset$, $\vee=\cup$, etc.).
The reason given that this is not a Boolean subalgebra is: The unit element of $A$ is $X$ which is not contained in $B$.
This seems fair.
Fast forward a few pages and we define a Boolean homomorphism (in the obvious way). Then the first example of a B homomorphism is:
Let $C$ be an arbitrary B alg. and let $p_0$ be an arbitrary element of $C$. Define $D = \{p: p\le p_0\}$ (where $p\le q$ means $p\wedge q = p$). Then let $0_D := 0_C$, $\wedge_D:=\wedge_C$, and $\vee_D := \vee_C$ but define $1_D := p_o$ and $\sim_D{p}:= p-p_0$. Then apparently $D$ is a Boolean subalgebra of $C$, and there is a natural homomorphism $\varphi:C\to D$ namely $p\mapsto p\wedge p_0$.
This seems correct too... but also appears to conflict with the counterexample given above. If the reason $\mathcal{P}(Y)$ is not a Boolean subalgebra of $\mathcal{P}(X)$ is simply that $1_A \not \in B$then it appears we are in the same problem with this boxed example since $1_C\mapsto 1_C\wedge p_0 \not \in D$ since after all $1_C \wedge p_0 = p_0$ and so it is not the case that $1\le p_0$.
So why is the former example not a Boolean subalgebra while the latter apparently is?
If $A$ is a subset of $B$, both carrying some structure, there is a difference between saying "there exists a homomorphism from $A$ to $B$" and "The inclusion map is a homomorphism from $A$ to $B$."
I guess that's what we're seeing here. If $A\subseteq B$, then there exists a boolean algebra homomorphism from $\mathcal{P}(A)\to\mathcal{P}(B)$ but the inclusion map $\mathcal{P}(A)\subseteq\mathcal{P}(B)$ isn't it, since the $1$ of $\mathcal{P}(A)$ is not the $1$ of $\mathcal{P}(B).$