Consider the following inverse system (edit: say category of multiplicative monoids or sets) \begin{equation} \cdots\rightarrow\mathbb{C}\xrightarrow{t\mapsto t^p}\mathbb{C}\rightarrow\cdots \end{equation} where $\mathbb{C}$ is complex numbers. The inverse limit is \begin{equation} \varprojlim_{t\mapsto t^p}\mathbb{C}=(a_0,a_1,\ldots, a_{i-1},a_i,\ldots) \end{equation} such that $a_i^p=a_{i-1}$. Intuitively this is $(a,a^{1/p},a^{1/p^2},\ldots)$ for some $a\in\mathbb{C}$, this can be multiplied with another tuple $(b,b^{1/p},b^{1/p^2},\ldots)$ to get $(ab,a^{1/p}b^{1/p},a^{1/p^2}b^{1/p^2},\ldots)$ giving it a structure of a multiplicative monoid.
In my opinion the universal property of inverse limit and algebraic closure of $\mathbb{C}$ should give $\mathbb{C}$ back. In other words the inverse limit described above should be isomorphic to $\mathbb{C}$ as a multiplicative monoid, i.e. instead of a tuple we should be just able to say $\mathbb{C}$.
Question: Is this true?
Sketch Proof [edited]: Let $\color{red}{\mathbb{C}}$ be a copy of $\mathbb{C}$ and a candidate for inverse limit and $a_0\in \color{red}{\mathbb{C}}$. Then consider a set theoretic map $a_0\mapsto a_i$ for $\mathbb{C}$ at $i$th position in the inverse system. This gives projection from $\color{red}{\mathbb{C}}$ to every $\mathbb{C}$ in the inverse system (See below).
\begin{equation} \cdots\underset{a_i}{\mathbb{C}}\rightarrow \underset{a_{i-1}}{\mathbb{C}}\rightarrow \cdots \end{equation}
We need to show that universal property is satisfied, for this consider the map $\varprojlim\mathbb{C}\rightarrow\color{red}{\mathbb{C}}$ as $(a_0,a_1,\ldots)\rightarrow a_0$, and the natural projections from $\varprojlim\mathbb{C}$ agree with natural projections from $\color{red}{\mathbb{C}}$ by construction. Now, the universal property of $\varprojlim\mathbb{C}$ gives the result as $\color{red} {\mathbb{C}}$. All maps are in {Sets}.
EDIT
The above holds for an algebraically closed field $k$ of char $p$, that is replace $k$ with $\mathbb{C}$ above, that is $\varprojlim_{t\mapsto t^p}k=k$.
Intuitively the tuple $(a_0=a,a_1,a_2,\ldots)$ in the inverse limit is just $(a,a^{1/p},a^{1/p^2},\ldots)$.
The question arises from the fact that there are two tuples corresponding to $1\in\mathbb{C}$. First the tuple $(1,1,1,\ldots)$ and second, the tuple $(1,\zeta_p,\zeta_{p^2},\ldots)$ where $\zeta_d$ is the $d$th root of unity. It seems to me that the sketch proof takes care of this by making a choice of a tuple.
One can consider $\mathbb{C}$ as a monoid with multiplication as group operation and $t\mapsto t^p$ as the homomorphism, or simply work with group $\mathbb{C}^\times$. In this case there is a clear projection map
\begin{equation} (ab,a^{1/p}b^{1/p},a^{1/p^2}b^{1/p^2},\ldots)\mapsto ab \end{equation} respecting the multiplication.
no this is not true. first, let us just look at everything as sets.as you noted in (3) the p-th power map is not injective over $\mathbb{C}$ so we do not have a canonical isomorphism between the inverse limit and $\mathbb{C}$. note that in field of char p case p-th power map is injective.
I think you still can find an isomorphism because these are sets and are determined only by their cardinality. but $\mathbb{C}$ has other structures as well. if you look at the multiplicative structure the p-th power map must be an isomorphism on the inverse limit so the inverse limit can not be isomorphic to $\mathbb{C}$ as monoids. and I don't know whether you can define a good additive structure on the inverse limit or not.
in fact know that I think about it we have an interesting ring structure on the inverse limit. if you forget about the topology $\mathbb{C}$ then it is the same as $C_p$: the completion of algebraic closure of $\mathbb{Q}_p$.so as a monoid$\varprojlim_{t\mapsto t^p}\mathbb{C}$ is the same as the monoid of tilt of $C_p$ which is an algebraicly closed field in char p. and this is another reason that the the multiplicative structure on $\mathbb{C}$ and inverse limit are very different from each other