I realize that the traditional definition for an order-embedding f is that
$a_1 \sqsubseteq a_2 \iff f(a_1) \sqsubseteq f(a_2)$
However, is it also fair to say that if $(A, \sqsubseteq)$ is a lattice, a definition for meet and this partial-order uniquely define one another:
$a_1 \sqsubseteq a_2 \iff a_1 \sqcap a_2 = a_1$
And so an alternative/equivalent way of saying f is an order-embedding might be:
$f(a_1 \sqcap a_2) = f(a_1) \sqcap f(a_2)$
Thank you! Any advice would be helpful.
Embeddings are functions from one structure to another which preserve the structure (and do no add additional structure to the image either).
For partially ordered sets this means, as you suggest that $a\sqsubseteq b\iff f(a)\sqsubseteq f(b)$. Lattices have more structure, so we can expect embeddings of lattices to preserve that structure. But note that this requires additional and explicit statements.
Consider the lattices $\mathcal P(X)$ and $\mathcal P(Y)$ ordered by inclusion, where $X=\{1,2\}$ and $Y=\{1,2,3,4\}$. Now define $f\colon\mathcal P(X)\to\mathcal P(Y)$: $$\begin{align} &f(\varnothing)=\varnothing\\ &f(\{1\})=\{1,4\}\\ &f(\{2\})=\{2,4\}\\ &f(\{1,2\})=\{1,2,3,4\}\end{align}$$
It is easy to see why $f$ is an order embedding, but the lattice structure is all skewered. $\{1\}\cap\{2\}=\varnothing$ but $f(\{1\})\cap f(\{2\})=\{4\}\neq f(\varnothing)$.
So while we can define a lattice structure on $\operatorname{Range}(f)$, this is not the structure inherited from the structure on $\mathcal P(Y)$, so in some sense it's cheating.
To sum up remember the motto, if you want more, assume more.