If $f:[0,t]\to[0,t]$ is continuous and increasing, why can we conclude $f(0)=0$ and $f(t)=t$?

Question

Let $t\ge0$ and $f:[0,t]\to[0,t]$ be continuous and (strictly) increasing. Why can we conclude $f(0)=0$ and $f(t)=t$?

I've tried the obvious thing: Let $\varepsilon>0$. Since $f$ is increasing and right-continuous at $0$, there is a $\delta>0$ with $$0\le f(s)-f(0)<\varepsilon\;\;\;\text{for all }s\in[0,\delta).\tag1$$ In $(1)$, I've used all the information we've got, but I don't see why this allows to conclude $f(0)<\varepsilon$ ...

Answer 1

With those conditions, we can't conclude that. For instance, take $t=1$ and $f(x)=\frac12x+\frac14$. Then $f$ is continuous and strictly increasing, but $f(0)=\frac14$ and $f(1)=\frac34$.

Seeing the new information, the exact quote is as follows (emphasis mine):

Let $\Lambda$ denote the class of strictly increasing, continuous mappings of $[0,1]$ onto itself. If $\lambda\in\Lambda$, then $\lambda0 = 0$ and $\lambda1 = 1$.

The word "onto" here means that we only consider the mappings which are surjective. And a surjective, strictly increasing, continuous function $f:[0,t]\to[0,t]$ certainly has $f(0) = 0$ and $f(t) = t$.