Is it correct to say L is RE if I can map reduction from LH to L?
Let's clear things a bit, since there are many equivalent/similar definitions that could lead to misunderstandings.


You can show that a language $L$ is recursive by constructing a recursive function $\chi_L$ (or a Turing machine or any other equivalent computation model) that decides it, i.e.
$$\chi_L (x) = \begin{cases} 1 &amp; \quad ; x \in L \\ 0 &amp; \quad ; \text{otw}. \end{cases}$$
notice that $\chi_L$ must be defined for all inputs.
You can show a that language $L$ is not recursive, by finding a "Turing reduction" from a non-recursive language to it. This is probably what what you mean by reduction, and it is defined as follows:


  We say that a language $A$ is Turing-reducible to a language $B$, written $A \le_T B$, if we can construct a recursive function (or Turing machine) that decides $A$ by assumption that there is such function for $B$.


As you see by definition, Turing reduction somehow "transfers recursiveness" from $B$ to $A$.


  For any $A$ and $B$ such that $A \leq_T B$, if $B$ is recursive, then so is $A$.


Hence if we already know that some $A$ is not recursive, then finding a reduction $A \leq_T B$ for any $B$ makes it non-recursive too.
You can show that a language $L$ is RE, by constructiong a recursive function $\varphi_L$ (or Turing machine) that $Dom(\varphi_L) = L$ (or halts on/accepts it), for example
$$\varphi_L(x) = \begin{cases} 1 &amp; \quad ; x \in L \\ \uparrow &amp; \quad ; \text{otw.} \end{cases}$$

where $\uparrow$ means "not defined" (or "does not halt"). There are other definitions of RE-ness, like its namesake "enumerability", which you can easily observe that are equivalent to this one.
But in showing a language $L$ is not RE, Turing reduction does not help, since it does not necessarily transfer RE-ness. For example observe that $\overline{L_{HALT}} \leq_T L_{HALT}$ (indeed any language is Turing-reducible to it's complement and vice versa), but we know that $L_{HALT}$ is RE, while $\overline{L_{HALT}}$ is not.

But there are other kinds of stronger reduction that do transfer RE-ness. One such reduction is called "many-one redicibility":


  We say that a language $A$ is many-one-reducible to a language $B$, written $A \leq_m B$, if there is a total recursive function $f$ such that for any input $x$
  $$x \in A \iff f(x) \in B$$


This is a stronger reduction in the sense that $A \leq_m B$ implies $A \leq_T B$ (and not necessarily vice versa). So, like Turing reduction, it transfers recursiveness. We also have


  For any $A$ and $B$ such that $A \leq_m B$, if $B$ is RE, then so is $A$.


To see how this is true, just take $\varphi_A(x) = \varphi_B(f(x))$.

Hence the correct method to show a language $B$ is not RE, is to find a many-one reduction $A \leq_m B$ for some non-RE language $A$. Notice that the above example does not work here, because $\overline{L_{HALT}} \nleq_m L_{HALT}$.


For further reading, see Computability Theory by S. Barry Cooper Pt. I, Ch. 7.