Is $K[X]\hookrightarrow K[[X]]$ an epimorphism? Let $K$ be a field and $X$ an indeterminate. 

Is the natural monomorphism $K[X]\hookrightarrow K[[X]]$ an epimorphism?

By epimorphism I mean epimorphism in the category of commutative rings.
EDIT. The question can be spelled out as follows: Are there distinct morphisms from $K[[X]]$ to some commutative ring $A$ which coincide on $K[X]$?
 A: The map from a Noetherian local ring to its completion is faithfully flat, and a faithfully flat epimorphism is an isomorphism. It follows from this that the map from the localization of $K[X]$ at $(X)$ to $K[[X]]$ is not epi. 
A: Let $B$ be a transcendence basis for the field $K((X))$ over $K(X)$ and let $L$ be an algebraic closure of $K((X))$.  Note that any injection from $B$ to itself can be extended to an endomorphism of $L$ over $K(X)$.  There are many such injections, since $B$ is uncountable (see below for a proof).  Restricting these endomorphisms to $K((X))$ gives many different homomorphisms $K((X))\to L$ which all agree on $K[X]$.  Restricting these homomorphisms to $K[[X]]$, we get many different homomorphisms $K[[X]]\to L$ which agree on $K[X]$.  (They are still different since a homomorphism on $K((X))$ is determined by its restriction to $K[[X]]$.)
[There are many other ways you could frame this same idea; for instance you could set it up more like uSir470888's answer.  The point is that it's really easy to define homomorphisms into an algebraically closed field.  In particular, once you're mapping into an algebraically closed field you can forget about whatever rigidity $K[[X]]$ might have had over $K[X]$ and map transcendental elements of $K[[X]]$ to any algebraically independent elements you want.]

Here is a proof that $K((X))$ always has uncountable transcendence degree over $K(X)$.  First, we prove a lemma:

Lemma: Let $k$ be a field, let $K$ be a field extension of $k$, and suppose $S\subset k((X))$ is algebraically independent over $k(X)$.  Then $S$ is algebraically independent over $K(X)$, as a subset of $K((X))$.
Proof: Suppose some elements $s_1,\dots,s_n \in S$ satisfy $h(s_1,\dots,s_n)=0$, where $h$ is a nonzero polynomial with coefficients in $K[X]$.  For each $m\in\mathbb{Z}$, the $X^m$ term of $h(s_1,\dots,s_n)$ can be written as a polynomial in the coefficients of the $s_i$ (which are certain elements of $k$) and the coefficients of the coefficients of $h$ (each coefficient of $h$ is an element of $K[X]$, and the coefficients of the coefficients of $h$ are elements of $K$).  Moreover, these polynomials are actually linear in the coefficients of the coefficients of $h$.  So to say that $h(s_1,\dots,s_n)=0$ is to say that the coefficients of the coefficients of $h$ satisfy a certain infinite list of linear equations with coefficients in $k$.  But if a system of linear equations in finitely many variables with coefficients in a field $k$ has a nonzero solution in an extension of $k$, it has a nonzero solution in $k$ (this is essentially the fact that the rank of a matrix cannot change if you extend the base field).  This means that we can replace the coefficients of $h$ by polynomials with coefficients in $k$ (which are not all zero) and $h(s_1,\dots,s_n)=0$ will still be true.  This contradicts the assumption that $S$ is algebraically independent over $k(X)$.

Now let $K$ be any field, and let $k$ be any countable subfield of $K$.  Since $k(X)$ is countable and $k((X))$ is uncountable, there exists an uncountable subset $S\subset k((X))$ which is algebraically independent over $k(X)$.  By the Lemma, $S$ is also algebraically independent over $K(X)$ as a subset of $K((X))$.
A: [Still checking if $g_1$ and $g_2$ are well defined. Let me know if you find something that doesn't. I changed the target ring to a larger one because I needed space for values on algebraic elements once I force a definition on a transcendental one.]
Let $R$ be the ring of power series on $X$ that are algebraic over the ring of fractions over $K$. This is $R$ consists of all power series $h$ such that there is a polynomial $p(Y)=a_0+a_1Y+...+a_nY^n$ with $a_i$ rational functions over $K$ with denominators that don't vanish at $X=0$, such that $p(h)=0$. 
We need to construct two different homomorphisms $g_1,g_2$ from $K[[X]]$ to $R$ such that they coincide on the polynomials. That way $g_1\circ i=g_2\circ i$ for the inclusion $i:K[X]\to K[[X]]$ but $g_1\neq g_2$. Showing that $i$ is not an epimorphism.
Define $g_i(s)=s$ for $s\in R \subset K[[X]]$, $i=1,2$. 
Now define $g_1(e^X)=1$ and $g_2(e^X)=\sum_{n=0}^{\infty}X^n=(1-X)^{-1}$. In particular $g_1(e^X)=1\neq (1-X)^{-1}=g_2(e^X)$. This defines $g_1,g_2$ on the sub-ring of $K[[X]]$ generated by $R$ and powers of $e^{nX}$ for $n\in\mathbb{Z}$, which are the elements of the form $\sum_{n=-N}^{N}e^{nX}a_n$ for $a_n\in R$. Observe how the transcendence of $e^X$ implies that each element of this sub-ring has a unique expression in this form. Their images by $g_1,g_2$ are $\sum_{n=-N}^{N}a_n$ and $\sum_{n=-N}^{N}(1-X)^{-n}a_n$ respectively.
We have $g_1,g_2$ defined in a larger sub-ring $R_{1}$. We still need to define them appropriately on the elements of $K[[X]]$ that are algebraic over $R_1$.
Assume $Y\in K[[X]]$ is algebraic over $R_{1}$. Then there is a polynomial $p(Z)=a_0+a_1Z+...+a_nZ^n$ with coefficients $a_i\in R_1$ such that $p(Y)=0$.
We need to define $g_1,g_2$ on $Y$ to be a solution of $0=g_i(a_0)+g_i(a_1)Z+...+g_i(a_n)Z^n$, respectively. In $R$ we have solutions of $0=g_i(a_0)+g_i(a_1)Z+...+g_i(a_n)Z^n$. So, we can send $Y$ to one of those solutions. This extends $g_1,g_2$ to a larger sub-ring $R_2$ of elements of $K[[X]]$ algebraic over $R_1$.
We can again pick an element of $K[[X]]$ transcendental over $R_2$. Defining their images in $R$ to be $1$ or $(1-X)^{-1}$ we can extend $g_1,g_2$ again and again to homomorphisms on ever larger subrings of $K[[X]]$.  
The end results will be ring homomorphism that are different at $e^X$ but equal on the polynomials.
