Every single algebraic topology problem I've encountered is leaving me baffled. I just can't visualise what is going on and why.
Q. $f,g:X \to S^1$ such that $f(x) \neq -g(x)$ for all $x \in X$. Construct a homotopy between $f$ and $g$.
is the question. Here's the answer
The line segment in $\mathbb{C}$ joining $f$ to $g$ does not pass through $0$.(uh, why? does it matter? Is there a problem if it does? what restricts it to NOT pass $0$? I am so lost) The map
$$h(x,t)=\frac{(1-t)f(x)+tg(x)}{||(1-t)f(x)+tg(x)||}$$
is well-defined since $(1-t)f(x)+tg(x) \neq \in \mathbb{C}$ and defines a homotopy.
So again, I don't get what it's trying to say when it says "The line segment joining $\mathbb{C}$ joining $f$ to $g$ does not pass through $0$" I mean, what does this matter?
And also, my answer to this question was $h(x,t)=(1-t)f(x)+tg(x)$. Just simply that. Not something like $h(x,t)=\frac{(1-t)f(x)+tg(x)}{||(1-t)f(x)+tg(x)||}$(I mean, yeah I guess this works as a homotopy as well, but is there a problem to keep it to $h(x,t)=(1-t)f(x)+tg(x)$?Simpler?)
And finally, what did the condition $f(x) \neq -g(x)$ have to do in solving this question? I don't see it come into play anywhere(well, I'm sure it did but not anywhere obvious)
As usual, not enjoying algebraic topology but I'm at least trying my best not to hate it, so any help is well appreciated, thanks