problems with proving that f and g are homotopic.

Question

i need to give an example of 2 continuous functions $f,g: X \rightarrow Y$ which are not homotopic, with: $X = [0,1] \times [0,1]$ and $Y = [0,1] \cup [2,3]$

and i need to show how many homotopical classes of continuous functions $f:X \rightarrow Y$ there exists.

for 1: i think the functions $f(x,y) = x$ and $g(x,y) = x+2$ will maybe work, using these functions we can make a function $F:X\times I \rightarrow Y$ with $F(x,t) = x+2t$. I find it confusing though by how to check wheter F is continous or not. Any hints about that?

for 2: I think the trick here is to let a random continuous function $f:X\rightarrow Y$ be homotopic to the zero-function $f'$, so we get one equavalention class. Is this the right way to think or am i on the wrong track?

Hints/tips and tricks would be very apreciated

Kees

Answer 1

For #1: Be careful! Your "homotopy" $F$ doesn't have values in $Y$ (and I guess you meant $F(x,y,t) = \dots$). For example $F(0, 0, \frac{3}{4}) = \frac{3}{2} \not\in Y$. To show that the two maps you define are not homotopic (and they are not), I advise you use the facts that:

• $X \times I$ is path-connected;
• $Y$ has two path-connected components: $[0,1]$ and $[2,3]$;
• the image of a path-connected space is path-connected.

So any you can't possibly have a homotopy $F : X \times I \to Y$ such that $F(x,y,0) = f(x,y)$ and $F(x,y,1) = g(x,y)$ (because the image would lie in two distinct path components.

For #2: $X$ is contractible and $Y$ has two contractible path components. How many homotopy classes of functions $\{0\} \to \{0,1\}$ are there?