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**60**

votes

**9**answers

15k views

### solving f(f(x))=g(x)

This question is of course inspired by the question How to solve f(f(x))=cosx
and Joel David Hamkins' answer, which somehow gives a formal trick for solving equations of the form $f(f(x))=g(x)$ on a ...

**14**

votes

**6**answers

2k views

### Topological characterization of the closed interval $[0,1]$

This question is related to question 92206 "What properties make $[0, 1]$ a good candidate for defining fundamental groups?" but is not exactly equivalent in my opinion. It is even suggested in one ...

**21**

votes

**4**answers

1k views

### is f a polynomial provided that it is “partially” smooth?

Let $f$ be a $C^\infty$ function on $(c,d)$ ,and
let $O=\cup_{n\in \mathbb{Z}^+} (a_n,b_n)$ where $(a_n,b_n)$ are disjoint open interval in $(c,d)$ and $O$ is dense in $(c,d)$.
Suppose for each $n\in ...

**10**

votes

**2**answers

1k views

### Does Riemann map depend continuously on the domain?

I've always taken this for granted until recently:
In the simplest case, given Jordan curve $C \subseteq \mathbb{C}$ containing a neighborhood of $\bar{0}$ in its interior. Given parametrizations ...

**4**

votes

**2**answers

207 views

### An approximate infinite-dimensional fixed point theorem

Given $\epsilon > 0$ and $f : [0, 1]^{\omega} \rightarrow [0, 1]^{\omega}$, can we find $x$ such that $x \in \textrm{Conv}\left( \left\{f(y) : ||y - x||_{\infty} < \epsilon\right\}\right)$?
In ...

**2**

votes

**1**answer

203 views

### Partitions of an interval

This question asks about properties of functions which are "piecewise" polynomials. I would like to ask a specific question about the meaning of "piecewise" there.
Specifically, consider "partitions" ...

**0**

votes

**1**answer

66 views

### Can we build a continuous function from “fibers”/preimages defined over a topological base?

I am looking for some proof insight or literature references for a statement which, if it's actually true, is probably a pretty trivial thing. I hope the question has not been asked here before, and ...