Homomorphic image of modular lattice is modular? Let $L$ be modular lattice, $M$ be lattice and $f:L\to M$ be a homomorphism. I want to show $f(L)$ is a modular lattice..
We already know that homomorphic image of a lattice is a lattice. 
So we only want to show that if $f(a)\leq f(b)$ then $f(a) \vee (f(x)\wedge f(b))= (f(a)\vee f(x))\wedge f(b) $ for $a,b,x \in L$
Since L is modular $a\leq b$ implies $a \vee(x\wedge b)= (a\vee x)\wedge b$
My problem is:
I must begin with the assumption $f(a)\leq f(b)$ then show $f(a) \vee (f(x)\wedge f(b))= (f(a)\vee f(x))\wedge f(b) $ for which I need to use $a \vee(x\wedge b)= (a\vee x)\wedge b$
But $f(a)\leq f(b)$ need not necessarily imply $a\leq b$ since it is only homomorphism and not isomorphism.
 A: A lattice is modular if and only if it satisfies the modular law:
$$a\le b\implies a\lor(x\land b)=(a\lor x)\land b.\tag1$$
Equivalently, a lattice is modular if and only if it satisfies the modular identity:
$$(c\land b)\lor(x\land b)=[(c\land b)\lor x)\land b.\tag2$$
Although $(1)$ and $(2)$ and not equivalent in isolation, they are equivalent in the presence of the other axioms of lattice theory; namely, their equivalence follows from the fact that, in a lattice, $a\le b$ if and only if $a=c\land b$ for some $c$.
Now, since $(2)$ is an identity, it is preserved by homomorphisms; therefore, a homomorphic image of a modular lattice is a modular lattice.
P.S. Here is a straightforward presentation which avoids mentioning the "modular identity":
Given $a,b,x\in L$ with $f(a)\le f(b)$, we want to show that
$$f(a)\lor(f(x)\land f(b))=(f(a)\lor f(x))\land f(b).\tag3$$
We don't know if $a\le b$, so let $a_0=a\land b$. Then $a_0\le b$, so by the modular law we have
$$a_0\lor(x\land b)=(a_0\lor x)\land b.\tag4$$
Moreover, $f(a_0)=f(a\land b)=f(a)\land f(b)=f(a)$ since $f(a)\le f(b)$. Since $f$ is a homomorphism, and since $f(a_0)=f(a)$, we see that $(3)$ follows from $(4)$.
A: You can instead replace $b$ with $a\vee b$, since $f(a\vee b) =f(a) \vee f(b) =f(b) $. 
A: First assume that $a\le c$ in $L$ such that $ a\vee(b\wedge c)=(a\vee b)\wedge c$ holds.
Now $a\le c\implies a\wedge c=a\implies f(a\wedge c)=f(a)\implies f(a)\wedge f(c)=f(a)\implies f(a)\le f(c)$.
So basically $f$, being homomorphism, carries out the order relation from $L$ to $M$. Therefore you can sit on $L$, make a relation $a\le c$ and carry forward it to $M$ as $f(a)\le f(c)$. No real need to reverse back from $M$ to $L$.
A: let Let :  →  be a homomorphism from a modular lattice L into a lattice M.
Again let () =  ⊆ . Now let , ,  ∈ , then there exist , ,  ∈ , such that () = ,
() = , () = . Suppose  ≤ .
 ∨ ( ∧ ) = () ∨ (() ∧ ())
= () ∨ ( ∧ ) since is homomorphism
= ( ∨ ( ∧ )) since is homomorphism
= (( ∨ ) ∧ ) since L is modular lattice
= ( ∨ ) ∧ () since is homomorphism
= (() ∨ ()) ∧ () since is homomorphism
= ( ∨ ) ∧  .
Hence,  ∨ ( ∧ ) = ( ∨ ) ∧  .
Therefore, homomorphic image of modular lattice is
