# What is the algebraic structure of functions with fixed points?

So I just noticed that the set of functions with a fixed point $$f(x_0)=x_0,$$ are closed under composition $$(f*g)(x):=g(f(x)),$$ and with $e(x)=x$, the inverible functions even seem to form a (non-commutative) group.

Then if one chooses another point $x_1$ and restricts the set to the functions which also have $x_1$ as a fixed point, then it is again closed and so on.

If I have one parameterized point (i.e. a curve, or even a couple of those), then solving $f_t(x(t))=x(t)$ for the families $f_t$ should give me morphisms between the functions for different values of $t$.

Are there general considerations regarding this?

And is this somehow related to the characterization of points of a manifold via the ideal of functions which evaluate to $0$ that point?

Edit 1: Might be just a general property of homeomorphisms or something, although I don't associating picking out isolated fixed points with these kind of things.

Edit 2: I now see that this might relate a translation/transformation of points in the manifold to a transformation of the function algebra over that manifold. This has some features: If you take two points $y_1$ and $y_2$ and transformations along the curves $Y_1(t),Y_2(t)$ with $Y_1(0)=y_1, Y_1(1)=y_2$ and $Y_2(0): =y_2, Y_2(1)=y_1$ (they move into each other), then the fuction set with both fixed points $Y_1(t),Y_2(t)$ makes a loop as $\{Y_1(0),Y_2(0)\}=\{Y_1(1),Y_2(1)\}=\{y_1,y_2\}$. The particular form of the curves have an impact on how the function set looks in between.

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+1 interesting thing... –  draks ... Aug 9 '12 at 8:46
Picking a fixed point just corresponds to removing that point from the domain and range of your functions (when looking at the invertible functions that is). –  Tobias Kildetoft Aug 9 '12 at 8:49
Congratulations: you have discovered the category of pointed sets/spaces/manifolds/etc. –  Zhen Lin Aug 9 '12 at 9:35
@Tobias: But that's only true for in the category of sets, not in the category of topological space, for instance. Right? –  Rasmus Aug 9 '12 at 9:37
You are looking at the stabilizer subgroup for a group action on a set. –  i. m. soloveichik Sep 13 '12 at 15:11