Let $A \colon= K[[X_1,X_2,\ldots,X_{\infty}]]$ be the formal power series ring with infinitely many variables over a field $K$. We can represent it also by the following manner$\colon$ \begin{equation*} A = \underset{n \geq 1}{\varprojlim}\, K[[X_1,\ldots,X_n]]. \end{equation*} $A$ is complete with the unique maximal ideal ${\frak m}_A$ which is closed and denoted by ${\frak m}_A = \overline{(X_1,X_2,\ldots,X_{\infty})}$. For example, we have the following inclusion$\colon$ \begin{equation*} \Sigma_{n=0}^{n=\infty}\,X_n = X_1 + X_2 + \ldots \in A. \end{equation*} Define the $K$vector space $V$ by the following$\colon$ \begin{equation} V \colon= {\frak m}_A/{\frak m}^2_A = \underset{{\lambda} \in \Lambda}{\bigoplus} K\omega_{\lambda}. \end{equation}
closed as unclear what you're asking by abx, RP_, JanChristoph SchlagePuchta, Stefan Waldmann, Olivier Feb 13 '17 at 8:17
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1$\begingroup$ $X_1^2$ is in both $\mathfrak{m}_A$ and $\mathfrak{m}_A^2$ $\endgroup$ – Will Chen Feb 12 '17 at 16:16

4$\begingroup$ Please fix your post before people possibly try to give you some answer: 1/ The ring of power series in infinitely many variables is not the projective limit you believe it is; 2/ What do you mean when you write $X_\infty$ ? 3/ What is $\Lambda$? What is $\omega_\lambda$? 4/ No need to shout; we can read math, even without boldface characters. $\endgroup$ – ACL Feb 12 '17 at 17:31

$\begingroup$ @ACL: Ironically, the one thing that is "right" in the question is that the formal power series he wants to use likely is the same as his definition of $A$; see my answer below. But I agree that it is full of other confusion, so not completely clear what is being sought (though I have a guess). $\endgroup$ – nfdc23 Feb 12 '17 at 18:43

1$\begingroup$ To anyone who is downvoting: though the question may look weird and/or meaningless, there is actually a genuine question lurking in there with a real theory behind it (though it seems the OP is not aware of that, explaining some of the mild confusion in the formulation), so please refrain from downvoting. The "flaws" in this question are no different from questions about higher categories that have imprecision in the formulation, except that the topic of pseudocompact rings is not as trendy as higher categories (and so imprecision about it is not as widely accepted for informal purposes). $\endgroup$ – nfdc23 Feb 12 '17 at 20:08

$\begingroup$ OK, in view of the OP's comments to my answer, I retract my suggestion not to downvote. There is in fact a wellposed question with genuine answer fitting into a useful framework, as developed in the SGA3 reference in my answer, but the OP is insisting (mistakenly, I believe) that this is not in the direction he wants, so nothing productive is going to emerge from further discussion until the OP thinks more carefully about the precise meaning of what he is asking about and looks closely at that SGA3 reference. $\endgroup$ – nfdc23 Feb 12 '17 at 22:08
Setting aside the meaningless "$X_{\infty}$", your notion of formal power series is an instance of that notion in an arbitrary (in your case countably infinite) set of variables as in the theory of pseudocompact rings in SGA3, Exp. VII$_{\rm{B}}$. More specifically, the completion of $K[X_1,X_2,\dots]$ for the system of ideals $$I_{J,N}=(X_j:j\in J)^N + (X_i:i\not\in J)$$ for $N>0$ and finite subsets $J \subset \{1, 2, \dots\}$ is topologically the same as $A$ with inverse limit of maxadic topologies as in the question.
It seems that you are looking for a criterion to make a "topological change of variables". But the version of Nakayama's Lemma appropriate for local pseudocompact rings (adapting the version for formal power series in a finite set of variables) is not the one that you seem to have in mind (and have not stated).
One issue is that the square of the maximal ideal is never closed when the number of variables is infinite, so the right quotient to consider for a notion of "cotangent space at the origin" is really $V_0 := \mathfrak{m}_A/\overline{\mathfrak{m}_A^2}$. A second issue is that the notion of "basis" of $V_0$ appropriate to consider here is not a raw algebraic basis in the sense of a direct sum but rather should be a topological $K$basis for $V_0$ viewed as a profinite $K$module (in the sense of SGA3, Exp. VII$_{\rm{B}}$: inverse limit of finitedimensional $K$vector spaces, equipped with topology as such). It is a general fact that if $W$ is any profinite $K$vector space then it is topologically free in the sense that $K^S \simeq W$ for some set $S$, where $K^S = \varprojlim_{S_0 \subset S} K^{S_0}$ for $S_0$ varying through the finite subsets of $S$; the elements $\{w_s\}_{s \in S}$ in $W$ corresponding to the evident "factors" $K$ of $K^S$ constitute a "topological $K$basis" of such a $W$. This $K^S$ is the same as $\prod_{s \in S} K$ with the product topology for the discrete topology on each copy of $K$.
To sum up, I believe you are asking the wrong question for whatever purpose you may have in mind (but since no motivation was given, I can only guess). I conjecture that what you really want to ask is this: for a subset $\{y_{\lambda}\}_{\lambda \in \Lambda}$ of $\mathfrak{m}_A$ lifting a topological basis of $V_0$, is the unique continuous map $$K[\![Y_{\lambda}]\!] \rightarrow A$$ of pseudocompact $K$algebras satisying $Y_{\lambda} \mapsto y_{\lambda}$ an isomorphism?
The answer is affirmative, but before I say anything more about that I'd like to get clear confirmation from you that this is actually addressing the question you want; if not then I likely will not have anything useful to say (and likely nor will anyone else).

$\begingroup$ Dear NFDC23, great thanks for your kind answer. The notation $K[[Y_{\lambda}]]$ in your definition implies
the inductive limit" of all possible $K[[Y_{\lambda},...,Y_{\gamma}]]$, where $(Y_{\lambda},...,Y_{\gamma})$ is the finitely many subset of $V_0$. That is, we would like to represent $A$ as the
attachment" of each finitelymanyvariable formal power series ring. If this is the case, please teach me how can I show that the answer is affirmative. Many thanks. $\endgroup$ – Pierre MATSUMI Feb 12 '17 at 20:32 
$\begingroup$ Also I have one question. m_A is already closed. Then why is the square of m_A not closed? Also what I would like to do is that I can rewrite A as the inductive limit of finitelymanyvariable formal power series ring, so I would like to get rid of inverse limit topology. $\endgroup$ – Pierre MATSUMI Feb 12 '17 at 20:44

$\begingroup$ My point of view is that each finitely many variables consist of finitely many subset of V_0, but I believe that m_A/(m_A)^2 being a Kvector space makes sense. I am afraid that for the characteristic ch(K) = p > 0, such element as X_1 + (X_2)^p + (X_3)^{p^3} + (X_4)^{p^4} + ... disappears in V_0, although it will be a member of the generators of m_A. With these problems in mind, I am now proving that A can be written as the inductive limit of finitelymanyvariable formal power series ring over $K$ by using generators of m_A, i.e. ω_{λ}'s such that (...,ω_{λ},...) = m_A. $\endgroup$ – Pierre MATSUMI Feb 12 '17 at 20:44

$\begingroup$ After all, I think that the lifting of the topological bases of V_0 is not what I am thinking. That is, the L.H.S. must be union of finitelymanyvariable formal power series ring with variables running over all generators of m_A. So, we need continuously many variables. Also, we need the presentation of inductive limit. My point of view is at least we have generators from the Kv.s. m_A/(m_A)^2, and will add more generators to V_0 to make up whole m_A. If we denote by Γ the index set, we have the presentation as A = lim K[[ω_λ,...,ω_ζ]] with (λ,...,ζ) running over all finite subset of Γ. $\endgroup$ – Pierre MATSUMI Feb 12 '17 at 20:54

$\begingroup$ Since (i) you are confused about inductive vs. inverse limits, (ii) you say you are trying to get rid of inverse limit topologies but that is a serious technical error, (iii) your 3rd comment has serious mistakes, (iv) your 4th comment is most definitely wrong (either the 1st sentence is wrong or else you are trying to do something that makes no sense), I'm going to withdraw from the discussion here. I don't have time to explain away the many misconceptions you have. I gave you the precise reference in SGA3, and urge you to read it to clear up the many misconceptions in your comments. $\endgroup$ – nfdc23 Feb 12 '17 at 22:03