Let $E = E\_n$ be the $n$-th Morava $E$-theory with coefficient ring
$$
E\_\* = \mathbb{W}(\mathbb{F}\_{p^n})[\![u\_1,\ldots,u\_{n-1}]\!][u^{\pm 1}].
$$
 It is usual to consider the completed co-operations
$$
E^\vee\_\* E := \pi\_\*L\_{K(n)}(E \wedge E)
$$
rather than the 'ordinary' co-operations $E\_\*E$. The latter is shown by [Hovey](http://projecteuclid.org/euclid.hha/1139839552) (although this was a folklore result) to be isomorphic to $\operatorname{Hom}^c(\mathbb{G}\_n,E\_\*)$ the ring of continuous functions from the $n$-th Morava stabilizer group to $E\_\*$ under the $\mathfrak{m} = (p,u\_1,\ldots,u\_{n-1})$-adic topology. 


Let $L\_0$ be the $L$-completion functor (for example see Appendix A of [Hovey-Strickland](http://hopf.math.purdue.edu/Hovey-Strickland/kn.pdf)). It is known that $E^\vee\_\*E = L\_0(E\_\*E) = (E\_\*E)^\wedge\_{\mathfrak{m}}$. There is a map $\phi:E\_\*E \to E^\vee\_\*E$ which can be identified with the usual $\frak{m}$-adic completion map.



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Standard commutative algebra tells us that the kernel of this map is isomorphic to $\bigcap\_{n=0}^\infty \mathfrak{m}^n E\_\*E$, although it seems difficult to approach this problem in this way. 


Here is a closely related question. Let $E(n)$ be the Johnson-Wilson cohomology theory with coefficient ring
$$
E(n)\_\* = \mathbb{Z}\_{(p)}[v\_1,\ldots,v\_{n-1},v\_n^{\pm 1}].
$$
Now there is an isomorphism $L\_0(E(n)\_\*E(n)) = E(n)\_\*^\vee E(n) \simeq \operatorname{Hom}^c(\mathbb{G}\_n,\widehat{E(n)}\_\*)$, where 
$$\widehat{E(n)}\_\* \simeq \mathbb{Z}\_{(p)}[v\_1,\ldots,v\_{n-1},v\_n^{\pm 1}]^\wedge\_I$$
for $I = (p,v\_1,\ldots,v\_{n-1})$. In this case by work of [Johnson](https://www.jstor.org/stable/25098982?seq=1#page_scan_tab_contents) it is known that there is an injection of $E(n)\_\*E(n)$ into $\operatorname{Hom}^c(\mathbb{G}\_n \times\mathbb{G}\_n,A)$ where $A$ the ring of integers in an unramified degree $n$ extension of the p-adic numbers. It is known that the latter is $L$-complete, and since $E(n)\_\*E(n) \to L\_0(E(n)\_\*E(n))$ is initial amongst maps with $L$-complete target, in this case it follows that $E(n)\_\*E(n) \to E(n)^\vee\_\*E(n)$ is injective.



Completed and uncompleted operations for Morava $E$-theory
No, this map is not injective. 


To see this, put $W=\mathbb{W}(\mathbb{F}\_{p^n})$, which is a free module of finite rank over $\mathbb{Z}\_p$. It is standard that $\mathbb{Z}\_p\otimes\mathbb{Z}\_p$ contains a rational vector space of uncountable dimension, so the same is true of $W\otimes W$. (Here tensor products are implicitly taken over $\mathbb{Z}$ or $\mathbb{Z}\_{(p)}$; you get the same anwswer either way.) This rational vector space can only map trivially to the group $E^\vee\_\*E$. Thus, it will suffice to check that $E\_\*E$ contains a copy of $W\otimes W$. The left and right unit maps provide a map $W\otimes W\to E\_0E$. In the opposite direction, we can define a map $\phi:E\_0\to W$ sending $u\_i$ to $0$ for $0<i<n$. The resulting formal group law over $W$ has logarithm $\sum\_kx^{p^{nk}}/p^k$, so the coefficients of the FGL lie in $\mathbb{Q}\cap W=\mathbb{Z}\_{(p)}$. Because of this, the two resulting FGLs over $W\otimes W$ are the same. Thus, the standard universal property of $BP\_\*BP$ gives a ring map 
$$ E\_\*E = E\_\*\otimes\_{BP\_\*}BP\_\*BP\otimes\_{BP\_\*}E\_\* \to W\otimes W, $$ 
which is left inverse to our map in the opposite direction.


