I'm slightly unsure about how the theorem is presented to me in lecture...

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> The Brouwer fixed point theorem for $1$ dimension. Every continuous map $[0,1] \to [0,1]$ has at least one fixed point.
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Well, the original theorem states the same except for that we deal with closed $n-$balls defined by

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> $$\bar{B}^n=\{(x\_1,...,x\_2) \in \mathbb{R}^n:x\_1^2+...+x\_n^2 \leq1\}$$
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So, the definition of this ball should, in $1$ dimension, give me the closed interval $[-1,1]$ no? Since we square it and any reals that give values less than or equal to $1$ is in $\bar{B}^1$.

Doesn't this include the negative reals too? Why is my lecturer only considering the interval $[0,1]$ only? I suppose it works for $[0,1]$ if the theorem works for $[-1,1]$ but just curious as to why this is the case. I could ask my professor but he's away on holiday :/

Does anyone have any idea? Is this, say, a "classical" or "conventional" way to present this particular theorem or something?