The problem is does there exist a continuous surjective function from $(a,b]$ to $(a,b)$
I am really not sure how to prove it but I do not think that it is possible. As $f(b)$ has to equal something but the function has to get close to $a$ and also $b$.
Many thanks
James
Yes, there is a continuous, surjective function $f:(a, b]\to (a, b)$. Such an $f$ is given by, for instance, $$ f(x) = \frac{b-a}2e^{-x+a}\sin\left(\frac{1}{x-a}\right) + \frac{b+a}{2} $$ As $x$ increases (toward $b$), the $e^{-x+a}$ factor will flatten the first term out so that $f(x)$ comes close to $\frac{b+a}2$. As $x$ decreases (towards $a$), the exponential factor will come ever closer to $1$, and the sine factor wil oscillate faster and faster between $-1$ and $1$. Therefore that whole term will oscillate faster and faster, and each bottom and top of that oscillation will come closer and closer to $\frac{a-b}2$ and $\frac{b-a}{2}$, so the entore function oscillates faster and faster and each oscillation comes closer and closer to filling the entire interval $(a, b)$.