(B)I note that if a function is continuous and injective, then very often the function is monotonic, and thus strictly monotonic, this is nearly always true?
That is long as the domain and range are connected, and/or are real valued, closed and bound intervals, that this is correct, and that the counter-examples often pertain, to disjoint intervals and sets,such as: $$[0,3)\cup{4}\to R$$.
Are these counter-examples only related to non-connected degenerate cases such as these?. I presume, they don not apply to apply when the domain and range are singular continuous, closed and bounded real intervals, or connected sets, such as $$[0,2]\to[0,3]$$?
Best Answer
To prove it you note that if the function is not strictly monotone you have a set of arguments where the value "turns"(*), that is $a<b<c$, but $f(b)$ is strictly greater (or less) than both $f(a)$ and $f(c)$.
Now select $y$ such that $y<f(b)$ but larger than both $f(a)$ and $f(c)$. By the intermediate value theorem you have that $f$ takes the value $y$ both somewhere in $(a,b)$ and $(b,c)$ which contradicts the assumption that $f$ is injective.
Note that the fact that the interval is closed was never used.
Of course the opposite is true: if a function is strictly monotone it's injective - which means that for continuous functions on an interval strict monotonicity and injectivity is equivalent.
(*) This is "obvious", but as often these can be proven (with some effort). You assume that the function is non-monotone which means that there is $b>a$ with $f(b)>f(a)$ and $d>c$ with $f(d)<f(c)$. Now it's basically to consider the possibilities of the relation between these arguments and the corresponding values - it's a couple of cases one need to cover.
For example if $b>d>c>a$ we have that $0<f(b)-f(a) = (f(b)-f(d)) + (f(d)-f(c)) + (f(c)-f(a))$, but as $f(d)-f(c)$ is negative at least one of the other terms must be positive. Which means that we have the claimed "turn" (either $b$-$d$-$c$ or $d$-$c$-$a$)