Why must a map from $X$ to $S^{n}$ not be onto for it to be null homotopic 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?

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.
