If $X$ is a topological space, we let $\text{End}(X)$ be the collection of continuous functions $f:X\to X$. We say that $f,g\in \text{End}(X)$ *meet* if there is $x\in X$ with $f(x) = g(x)$. We say that $D\subseteq \text{End}(X)$ is a *cover* for $\text{End}(X)$ if for every $f\in \text{End}(X)$ there is $g\in D$ such that $f$ and $g$ meet.


For $\text{End}(\mathbb{R})$, a boring example of a cover is the collection of all constant functions. A more interesting example is the following countable cover: for $k\in\mathbb{Z}$ let $f\_k$ to be defined by $x \mapsto x+k$ for all $x\in\mathbb{R}$. Let $c\_0$ be the constant $0$ function. Then $$\{f\_k:k\in \mathbb{Z}\}\cup\{c\_0\}$$ is a cover for $\text{End}(\mathbb{R})$.


This motivates the following question: If $\kappa$ is an infinite cardinal and $X$ is a connected $T\_2$-space with $|X|=2^{\kappa}$, does $\text{End}(X)$ have a cover of cardinality $\kappa$? 



Continuous function covers in connected $T_2$-spaces
The answer is strong YES for connected spaces admitting a non-constant continuous function and NO in the opposite case.


1. If $X$ is a connected topological space that admits a non-constant continuous function
$\gamma:X\to\mathbb R$, then the family 
$$\{f\in C(X):f(X)=\{q\}\subset\mathbb Q\}\cup\{q+\gamma:q\in\mathbb Q\}$$is a countable cover for $C(X)$.
2. If $X$ is a connected space admitting no non-constant continuous function, then the family $C(X)$ consists of constant functions, has cardinality of continuum and each covering of $C(X)$ has cardinality of continuum.


**Remark.** Examples of connected Hausdorff spaces without non-constant continuous functions are well-known, take for example, the Golomb space. Or even a singleton for the trivial case.


**Added in Edit.** It turned out that I answered another question mixing $\mathrm{End}(X)$ with $C(X,\mathbb R)$. But for $\mathrm{End}(X)$ also there exists a counterexample: the [Cook continuum](http://matwbn.icm.edu.pl/ksiazki/fm/fm60/fm60123.pdf) $K$. It has countable weight but each continuous self-map $K\to K$ is constant. So each cover for $\mathrm{End}(K)$ has cardinality $\mathfrak c>\omega=w(K)$. 


Nonetheless, the question remains open for Peano continua: *Has the set $\mathrm{End}(X)$ a countable cover for each Peano continuum $X$ without the fixed point property?* The answer is not known even for compact connected Lie groups, for example the multiplicative group $S^3$ of quaternions of unit norm.


Let us observe that for $n\le 2$ the $n$-dimensional sphere $S^n$ has a countable cover. More precisely, for $n\in\{0,2\}$ the family $\{\mathrm{id},-\mathrm{id}\}$ is a 2-element cover of $S^n$. For $n=1$, fix any countable dense set $Y\subset S^1$ and observe that the family $\{\mathrm{id}\}\cup\{f\in\mathrm{End}(S^n):\exists y\in Y\;f(S^n)=\{y\}\}$ is a countable cover of $\mathrm{End}(S^n)$. So, the first unclear case is the 3-dimensional sphere $S^3$.


