This question was inspired by a post from Gian Maria Dall'Ara concerning ways to prove the Fundamental Theorem of Algebra. In particular, see this MO post, where he formulates the FTA as:

Every complex non-constant polynomial $p$ is surjective.

My own thoughts are offered below, but the main question here is as stated in the title:

Can one prove the Fundamental Theorem of Algebra using just fixed point theory?

[Edit: The writing below is left as an artifact from the original post, but need not be addressed in formulating an answer to the actual question posed above. Perhaps another MO user would be willing to rephrase it more succinctly and reformat using $\LaTeX$...]

OK, so if take out the subset of points from the domain where the derivative of the function is zero (critical points) and remove from the codomain the set of points formed from the image of the critical points, you are left with 2 sets of sets – one domain set without critical points and one with critical points as well as one codomain set without the images of the critical points, and one codomain set with images of critical points. Let A be the domain set without critical points and let B be the codomain set without images of critical points. Obviously, removing a bunch of points from the complex plane does not render the plane disconnected because you can always get from one point to another by going around the "holes" left by the removed points. So A and B are both connected but no longer simply connected. Critical points are where the derivative is zero or does not exist. The next connection is to use topological fixed point theory. If there is a function f on a space without fixed points, then there is a continuous retraction of that space to its boundary. So” no fixed points” implies there is a continuous retraction to the boundary. Contra-positively, “no continuous retraction” imply “no f without fixed points.” OK now for a little bit of trivial algebra. Fixed points have this form: f(z)=z take the derivative, and you get f'(z) =df/dz = dz/dz=1 so f'(z) =1 for a fixed point, or f'(z)-1=0 but for critical points f'(z)=0 by definition, so substituting we get 0-1=0 which is a contradiction, so the assumption of fixed points existing in the set of critical points is false. By fixed point theory then, there is a continuous retraction of the space to its boundary (This is kind of what you see in the Holographic Principle, Stoke's theorem, Green's Theorem, Divergence Theorem etc.)

Now why does a continuous retraction of a topological space to its boundary have to do with surjectivity? It means that for any transformation (endomorphism) of the space, there is a continuous retraction of that space to its boundary. The complex numbers C form a connected locally compact topological field. So associate the boundary with the domain C, and the interior with the range, and we get surjectivity of the set of critical points. From Category Theory, there are two aspects to inversions of maps generally - retractions and sections. If we have a map (f) from space A to space B, the composite morphism of the retraction r with the map gives the identity in the domain. A-------f--------->B B--------r---------->A where r * f = 1A (the identity element in domain A)

Applying these categorical concepts to the situation at hand we have. The function f is surjective because there is a continuous retraction for every point in the codomain back to the domain. OK, so the set of critical points has a continuous retraction and is therefore surjective. What about the remaining open set where the critical points have all been removed? How do we show that it is surjective with respect to the domain of the polynomial for which we are seeking roots? The fact that this set (of polynomial transformations) is compact (closed and bounded) and connected means that there could indeed be fixed points in that set. This would mean that the function would not be surjective when inclusive of the fixed points. Are we in trouble now because the whole purpose was to show that the map (complex polynomial) was surjective, and we just showed that it might not be? I don’t think so, because we just showed that when fixed points occur, the derivative gives a constant that is not zero. All constants except 0 are precluded from examination. Therefore, there are no fixed points in the codomain set which excluded the critical points. Let’s me go over this part a little more slowly. Now B may contain fixed points (especially when considering the base field of complex numbers which are compact and connected.) OK so consider fixed points in B. This is where f(z)=z in B. However, recall that B contained no critical points i.e. where f’(z)=0. This means there are no points in B where f(z)= constant because it contains no critical points. Why? Because when you integrate f’(z)=0 you get f(z)= constant. However, f(z) not being equal to a constant precludes any constant value of z in the entire complex domain. So excluding critical points amounts to excluding the entire domain C of complex numbers if we assume that there are fixed points in B. Therefore, B has no fixed points, and by the above topological fixed point theory, there is a continuous retraction from B to its boundary. If we now associate the boundary with A and the open map f to B, we have the surjection we were looking for in the retraction from B obtained from the fact there were no fixed points in B. It is kind of obvious, that if you remove a lot of points in the complex plane (e.g. in this case critical points), that you no longer have a simple connected domain because you have a plane full of holes. Therefore, once you have holes, you are no longer guaranteed fixed points anyway (Brower). Therefore in such a case, there can be continuous retractions to the boundaries provided that other topological features such as compactness prevail. The surjectivity of complex polynomials then follows from the continuous retraction to the boundary of the complex space.