Homomorphic image of modular lattice is modular?

Question

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.

Answer 1

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 $(4)$.

Answer 2

You can instead replace $b$ with $a\vee b$, since $f(a\vee b) =f(a) \vee f(b) =f(b) $.

Answer 3

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 $L$.

Answer 4

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