Two versions of invariance of domain theorem? While reading about Invariance of Domain Theorem, I noted that there are two common version of it, one saying that $f(U)$ is open, while another saying that $f$ is a homeomorphism (and sometimes both of these, like in the wikipedia link).
But I only saw proofs (1 and 2) for $f(U)$ being open, does this imply $f$ being an homeomorphism? And if it does, how? 
Regarding proof 1, I have two more questions, if possible:


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*Why does he say that it will be enough to prove that $f(0)$ is in the interior of $f(B^n)$? $\,\,\,$ i.e. why does theorem #5 imply theorem #2 (invariance of domain)? What if $0\notin U$, how can it still assist us?

*We know that $f^{-1}: f(B^n) \rightarrow B^n$ is continuous, but why can we extend it to a continuous $G:\mathbb{R}^n \rightarrow \mathbb{R}^n$? As far as I understand it, Tietze's extension theorem can be used on a function $h:A\rightarrow \mathbb{R}$ where $A$ is closed, but in here $f(B^n)$ is not closed and we are working with $\mathbb{R}^n$ which is not homemorphic to $\mathbb{R}$, so how comes this theorem can be used in here?
 A: 
But I only saw proofs for $f(U)$ being open, does this imply $f$ being an homeomorphism? And if it does, how?

Yes, as Daniel Fischer answered in the comments. Restricting the codomain of an injective map to its image yields a bijection. If $V$ is any open subset of $U$, then the pair $(V, f\rvert_V)$ also satisfies the hypothesis of the "open" version of the theorem, so $f(V)$ is open. This shows that $f^{-1}$ is continuous.

  
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*Why does he say that it will be enough to prove that $f(0)$ is in the interior of $f(B^n)$? $\,\,\,$ i.e. why does theorem #5 imply theorem #2 (invariance of domain)? What if $0\notin U$, how can it still assist us?
  

It seems that you may be a bit confused by the notational overload on $f$. Note that $f$ refers to distinct, unrelated maps in Theorems 2 and 5 (different domains, for one thing).
Here's a hint: To prove Theorem 2, it suffices to show that for each $x \in U$, we have $f(x) \in \operatorname{int} f(U)$. So for each $x \in U$, try to find a continuous injective map $g_x \colon B^n \to \mathbb R^n$ (note that $g_x$ satisfies the hypothesis of Theorem 5) that will help you do exactly that.


  
*We know that $f^{-1}: f(B^n) \rightarrow B^n$ is continuous, but why can we extend it to a continuous $G:\mathbb{R}^n \rightarrow \mathbb{R}^n$? As far as I understand it, Tietze's extension theorem can be used on a function $h:A\rightarrow \mathbb{R}$ where $A$ is closed, but in here $f(B^n)$ is not closed and we are working with $\mathbb{R}^n$ which is not homemorphic to $\mathbb{R}$, so how comes this theorem can be used in here?
  

The set $f(B^n)$ is actually closed. Perhaps you were under the impression as I was that $B^n$ refers to the open unit ball, but in fact, Tao defines $B^n$ as the closed unit ball at the outset (in the statement of Theorem 1). As $f$ is continuous, $f(B^n)$ is compact, hence closed.
We can then use the fact that a map into a product space is continuous iff each of its component maps is continuous. Break $f^{-1}$ into its component functions $(f^{-1})_i \colon f(B^n) \to \mathbb R$, extend by Tietze to $G_i \colon \mathbb R^n \to \mathbb R$, then take $G := (G_1, \dotsc, G_n)$.
