# Question regarding homomorphism

In James.E.Munkres There is a question that says prove that subspace $(a,b)$ of $\mathbb R$ is homeomorphic with $(0,1)$.

In the definition of homeomorphism it is given that Let $X$ and $Y$ be two topological spaces.

Let $f:X \rightarrow Y$.Now if both the function $f$ and it's inverse function are continuous, then the function $f$ is called homomorphism.

Now my question is shouldn't there be a function two prove if two sets are homeomorphic??? Any help would be appriciated. Thanks.

• "then the function is called a homeomorphism" and this is indeed the desired function to show that two topological spaces (sets with a topological structure) are homeomorphic. – Andres Mejia May 22 '17 at 12:37
• So I have to find the desired function?? – user426700 May 22 '17 at 12:38
• For your exercise, you just need to show a continuous function $f:(a,b)\rightarrow(0,1)$ whose inverse function is also continuous. Hint: don't look for smg too complicated ! – Evargalo May 22 '17 at 12:39
• Oh. OK. Thanks. – user426700 May 22 '17 at 12:41

Intuitively, every interval $(a, b)$ in $\mathbb{R}$ is indeed homeomorphic to the unit interval, $(0, 1)$ (notice only open intervals are homeomorphic to each other; $[a, b)$, for instance, is not homeomorphic to $(a, b)$). You can see this because you can either stretch or shrink (ie, "continuously deform") $(a, b)$ into the unit interval, and vice versa.
You rigorously prove that two objects are homeomorphic to each other by finding an explicit map, a homeomorphism, between the two objects, which is the bicontinuous bijection defined in your question. For instance, if I were to prove that $(2, 4)$ is homeomorphic to $(1, 0)$, one way to go about finding a homeomorphism is parametrizing $(2, 4)$. In other words, I need to find $f: \mathbb{R} \rightarrow \mathbb{R}$, $x \mapsto tx$, where $x \in (2, 4)$ and $tx \in (1, 0)$, where $0 \leq t \leq 1$ (this works in general for any homeomorphisms to a unit interval, circle, sphere, etc).
$$x = 2 + t(4 - 2) = 2 + 2t \implies t = \frac{x - 2}{2} \implies f(x) = \bigg(\frac{x - 2}{2}\bigg) x$$. You can easily verify that $f$ and its inverse are continuous (they are just polynomial/ rational functions!) so we have a homeomorphism. You can just use the same idea for $(a, b)$.