The problem is...
Suppose that $X = \lim_\leftarrow \{X_i, f_i\}_{i=1}^{\infty}$ is an inverse limit space so that there is an integer $N$ and for each $n \geq N$ the function $f_n$ is an onto homeomorphism. Then $X$ is homeomorphic to $X_N$.
Can't we just define a map $f: X \rightarrow X_N$ to be $f(P) = P_N$, where $P=\{P_i\}_{i=1}^{\infty} \in X$? Won't this be a homeomorphism? Why do we need an integer $N$ that satisfies such conditions? Where do I use that the $f_n$'s are onto homeomorphisms?
Throughout I assume that we are talking about the concrete topological realization of the inverse limit, namely as a special subset of the product space.
So you claim that the natural projection is a homeomorphism. Obviously that needs a proof. So lets construct the inverse $g:X_N\to X$. Note that $g(x)$ is a sequence. Denote by $g(x)_k$ its $k$-th coordinate. Define:
$$g(x)_{N+k}=f_{N+k-1}^{-1}\circ\cdots\circ f_N^{-1}(x)$$ $$\vdots$$ $$g(x)_{N+2}=f_{N+1}^{-1}(f_N^{-1}(x))$$ $$g(x)_{N+1}=f_N^{-1}(x)$$ $$g(x)_N=x$$ $$g(x)_{N-1}=f_{N-1}(x)$$ $$g(x)_{N-2}=f_{N-2}(f_{N-1}(x))$$ $$\vdots$$ $$g(x)_{N-k}=f_{N-k}\circ\cdots\circ f_{N-1}(x)$$
Can you see why I've defined $g$ like this? $f_i$ obviously have to be homeomorphisms for $i\geq N$ in order for $g$ to be well defined and continuous.
Note that this is crutial. Without these assumptions the natural projection need not be a homeomorphism and often is not.
Side note: A function is not just formula. You seem to think that the natural projection does not depend on $f_i$ because its formula doesn't. This is wrong. It very much does, because the domain does. And so properties of the natural projection depend on $f_i$. Similarly the projection onto first coordinate $\mathbb{R}^2\to\mathbb{R}$ is not a homeomorphism but the same projection $\mathbb{R}\times\{1\}\to\mathbb{R}$ is. A function is made up of three things: formula, domain and codomain. And this is a great example of that fact.