It would really be $f|_{I^n}$ a homeomorphism onto $I^n$ and $f (I^n)$?

Question

I am studying an article that deals with the proof of invariance of the domain theorem. In the article the following statement is made:

Since $I^n$ is a compact space, the map $f|_{I^n}$ is a homeomorphism from $I^n$ to $f(I^n)$.

In this part of the article,

• $I^n$ is the n-dimensional cube $[-a,a]^n$ and $a>0$,
• $f:U\to \mathbb{R}^n$ is an injective continuous map,
• $U\subset \mathbb{R}^n$ is open,
• $I^n\subset U$.

But this statement is not clear to me. It is not clear to me what the topology or the open sets of the compact $I^n$" are. Likewise, it is not clear who the open sets or topology of $f(I^n)$ are.

My question. How can I prove this statement and make it clear who are the topologies of $I^n$ and $f(I^n)$ that makes the constraint $f|_{I^n}$ a homeomorphism onto $I^n$ and $f (I^n)$?

Answer 1

Here, $I^n$ and $f(I^n)$ are subspaces of $\mathbb{R}^n$. That is, their topology is the topology induced by $\mathbb{R}^n$. And $f|_{I^n}$ is a homeomorphism because $f$ is continuous and injective and $I^n$ is compact.