Example of non-homotopic functions I  start to study homotopy theory and I've trouble with following "proof":
"If $f_0$, $f_1$ are continous functions $X \rightarrow Y$, then one can consider the continous function  $F=(1-t)\cdot f_0+ t\cdot f_1 :X\times [0,1] \rightarrow Y$. 
Hence any two continous functions are homotopic." 
Please, give a counterexample and find a mistake.  
 A: If $Y$ is the (vector space of) complex numbers, your argument is correct as stated.
If $Y$ is a proper subset of the complex numbers, then the mistake is this:
$$
F(x, t) = (1 - t) \cdot f_{0}(x) + t \cdot f_{1}(x)
$$
need not be an element of $Y$ even if $f_{0}$ and $f_{1}$ take values in $Y$. Geometrically, for each $x$ the mapping $t \mapsto F(x, t)$ parametrizes the segment joining $f_{0}(x)$ and $f_{1}(x)$. If this segment is not contained in $Y$, then $F$ does not take values in $Y$.
For example, let $X = [0, 2\pi]$, $Y = \mathbf{C}\setminus\{0\}$ the set of non-zero complex numbers, and consider the continuous mappings $f_{0}$, $f_{1}:X \to Y$ defined by
$$
f_{0}(x) = 1,\qquad
f_{1}(x) = e^{ix}.
$$
The "straight-line homotopy" $F$ does not map $X \times [0,1]$ into $Y$. I leave to you the easy and instructive exercise of finding the pair $(x, t)$ with $0 \leq x \leq 2\pi$ and $0 \leq t \leq 1$ such that $F(x, t) \not\in Y$.
Fine point: The maps $f_{0}$ and $f_{1}$ above are actually homotopic in $Y$ unless you impose additional conditions, e.g., the homotopy fixes endpoints of paths. (The preceding paragraph merely shows that the mapping $F$ is not a homotopy from $f_{0}$ to $f_{1}$ considered as $Y$-valued mappings.)
A more blunt example is to take $Y = \{0, 1\}$ (a set of two complex numbers) and $X = \{*\}$ (a singleton), and to define $f_{0}(*) = 0$, $f_{1}(*) = 1$. These maps are certainly not homotopic in $Y$, and it's obvious why the straight-line homotopy $F$ "doesn't work".
