Gamelin "Topology" has an exercise 3.3.1 to prove that if $n \geq 2$ then $S^{n}$ is simply connected. Then as a hint he suggests: show that every loop in $S^{n}$ is homotopic to a loop that does not cover all of $S^{n}$.
Similarly, Armstrong "Basic Topology" has in Exercise 5 (p. 91) Let $f: X \rightarrow S^{n}$: be a map which is not onto. Prove that $f$ is null homotopic.
I would appreciate help understanding – especially on an intuitive basis – what goes wrong if $f$ is not onto or why you have to exclude loops that cover all of $S^{n}$.
I know you can extend straight-line homotopy to two functions when $f,g:X\rightarrow S^{n}$ never give a pair of antipodal points. But,again, although the normalized form $F(s,t) = \frac{(1-t)f(x)-tg(x)}{\|(1-t)g(x)-tg(x)\|}$makes this assertion clear, I still don't have an intuitive sense as to why it is necessary.
Lastly, adding to my confusion, is the theorem that in a convex subset of $\mathbb{R^{n}}$, any two paths with endpoints fixed are homotopic.
ADDENDUM Maybe I should also state my problem in simple terms. Gamelin, in the solution to his problem above says: If $\gamma$ is a loop based at the south pole that does not pass through the north pole, we can homotopy $\gamma$ to a point by pulling continuously to the south pole along circles of longitude.
So why does he impose that the loop not pass through the north pole?
Best Answer
If the image of $f$ doesn't contain a point $p\in S^n$, then $f$ is a composition $f:X \to S^n\setminus \{p\} \to S^n$. Since $S^n\setminus\{p\} = \mathbb{R}^n$ is contractible, the map $X \to S^n\setminus\{p\}$ and thus $f$ must be null-homtopic.