In group theory, is it true that $f(X \vee Y) = f(X) \vee f(Y)$? ($\vee$ denotes join).
Let $G$ and $H$ denote Abelian groups, $X$ and $Y$ denote subalgebras of $G$, and let $f : G \rightarrow H$ denote a homomorphism. Then:
$$f(X \vee Y) = f(X+Y) = f(X)+f(Y) = f(X) \vee f(Y)$$
So $f(X \vee Y) = f(X) \vee f(Y)$.
Now suppose we relax our conditions, so that $G$ and $H$ are no longer assumed Abelian.

Question. Is it still provable that $f(X \vee Y) = f(X) \vee f(Y)$?

 A: That we have $f(X\vee Y)\subseteq f(X)\vee f(Y)$ should be clear (the image of a product of elements in $X$ or $Y$ is a product of elements in $f(X)$ or $f(Y)$).
For the other direction, let $g\in f(X)\vee f(Y)$. Then $g=g_1\cdot\cdots\cdot g_n$ where $g_i\in f(X)$ or $g_i\in f(Y)$. Thus each $g_i$ is of the form $f(x_i)$ where $x_i\in X$ or $x_i\in Y$. Thus we can rewrite $g=f(x_1\cdot\cdots\cdot x_n)$. Thus $g\in f(X\vee Y)$.
This approach can be generalized for arbitrary collections of subgroups since it didn't matter that $\{X,Y\}$ was a finite collection. The only thing that was important was that elements were a finite product of other elements. Thus for any collection of subgroups $\cal S$ of $G$, we have that $f(\bigvee_{S\in\cal S} S)=\bigvee_{S\in\cal S}f(S)$.
The dual question is however false. In general, we do not have that $f(X\wedge Y)=f(X)\wedge f(Y)$ for arbitrary subgroups $X$ and $Y$, specifically the inclusion $f(X)\wedge f(Y)\leq f(X\wedge Y)$ need not hold (the other inclusion always holds). I was having trouble finding a counterexample, so I asked here where Marcin Łoś gave a counterexample in the comments. Thus in general, a group homomorphism does not induce a lattice homomorphism between the subgroup lattices.
However, if $f$ is a quotient map (or surjective by the 1st isomorphism theorem), the 4th isomorphism theorem does give us some relation between subgroup intersections.
A: Here is how I see it (without reference to "messy" product representations).
If we partially order the lattice of subgroups of $G$ by inclusion, then $H \vee K = \langle H,K\rangle$, the subgroup generated by the set $H \cup K$. By definition this is the smallest subgroup (via our partial order) of $G$ that contains both $H$ and $K$.
Now for any function $f:A \to B$, if $X \subseteq Y \subseteq A$, we certainly have: $f(X) \subseteq f(Y)$.
Homomorphisms are functions, so we have $f(H) \subseteq f(\langle H,K\rangle)$, and similarly for $f(K)$ (since $H,K \subseteq \langle H,K\rangle$).
This shows that $\langle f(H),f(K)\rangle \subseteq f(\langle H,K\rangle)$, by minimality (there is a "hidden appeal" here to the fact $f$ is a homomorphism, so we may conclude $f(\langle H,K\rangle)$ is a group, as our property of "smallest subgroup that contains the generating set" only applies to elements of the subgroup lattice, not arbitrary subsets of $f(G)$).
Now suppose $L$ is ANY subgroup of $f(G)$ containing $f(H)$ and $f(K)$. It follows that $f^{-1}(L)$ is a subgroup of $G$ containing $H$ and $K$ (this is again where we invoke the fact that $f$ is a homomorphism), so:
$\langle H,K\rangle \subseteq f^{-1}(L)$
Thus $f(\langle H,K\rangle) \subseteq f(f^{-1}(L)) = L$.
Since $\langle f(H),f(K)\rangle$ is such a subgroup, we have: $f(\langle H,K\rangle) \subseteq \langle f(H),f(K)\rangle$, and equality is established.
