@0celo7 you are literally saying that if the $f$ in $f : \mathbb{R}^n \rightarrow \mathbb{R}^n | \vec{x} \mapsto f(\vec{x})$ is such that $f(\vec{x}) = 0$ then this means $\mathbb{R}^n = {0}$
@0celo7 $\{0\}$, the point $0$ in \mathbb{R}^2 is completely different from the point $0$ in \mathbb{R}^3 and neither are elements of a zero-dimensional manifold, they are just points in a space of $n$ dimensions
@bolbteppa But if you are contracting $\mathbb{R}^n$ itself (to a point), how does it make sense to talk about the origin like it's still in $\mathbb{R}^n$?
@KyleKanos , @dmckee , @DavidZ there are too many posts with no accepted answers, but with comments that in fact answer the question. When I see such comments I send to the commentator an advice to transform the comment into an answer. But I insist that you proceed the same way, i.e. send messages to users to transform their solving comments into an answers. I pass sometimes over apparently unanswered questions, and need to read a whole correspondence for finding out if it is answered or not.
> It remains a mystery how black holes could have grown so huge in such a relatively brief time after the dawn of the universe, researchers say.
> "This is quite surprising because it presents serious challenges to theories of black hole growth in the early universe," said lead study author Xue-Bing Wu, an astrophysicist at Peking University in Beijing.
@0celo7 contractible is a property of functions it does not affect the actual sets, that's kind of all I can say, you use functions to analyze contractibility, you can't actually deform sets the intuitive way you're thinking so you have to use functions to say these things and use loose language.
I would be very interested if someone would solve this problem. I understand it but I don't know the matrixes $\gamma$. And, moreover, what is $\gamma_5$.
