- #1

cianfa72

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- TL;DR Summary
- Homemorphism between a linear subspace of ##\mathbb R^8## and ##\mathbb R^4##

Hi,

consider the Euclidean space ##\mathbb R^8## and the projection map ##\pi## over the first 4 coordinates, i.e. ##\pi : \mathbb R^8 \rightarrow \mathbb R^4##.

I would show that the restriction of ##\pi## to the linear subspace ##A## (endowed with the subspace topology from ##\mathbb R^8##) is homeomorphism with ##\mathbb R^4##. ##A## is defined by

$$x_5 = x_1, x_6=x_2, x_7=x_3, x_8 = x_4$$

##\pi## is continuous and open, its restriction to ##A## is continuous as well. What about openess in the subspace topology ?

My idea is to show that projection's restriction is open as well using open balls as topology basis in ##\mathbb R^8##. The intersection of an open ball with ##A## can be either empty or a set such that its projection is open in ##\mathbb R^4##. Therefore we have an open continuous bijection between ##A## and ##\mathbb R^4## hence they are homemorphic.

Does it make sense ? Thanks.

consider the Euclidean space ##\mathbb R^8## and the projection map ##\pi## over the first 4 coordinates, i.e. ##\pi : \mathbb R^8 \rightarrow \mathbb R^4##.

I would show that the restriction of ##\pi## to the linear subspace ##A## (endowed with the subspace topology from ##\mathbb R^8##) is homeomorphism with ##\mathbb R^4##. ##A## is defined by

$$x_5 = x_1, x_6=x_2, x_7=x_3, x_8 = x_4$$

##\pi## is continuous and open, its restriction to ##A## is continuous as well. What about openess in the subspace topology ?

My idea is to show that projection's restriction is open as well using open balls as topology basis in ##\mathbb R^8##. The intersection of an open ball with ##A## can be either empty or a set such that its projection is open in ##\mathbb R^4##. Therefore we have an open continuous bijection between ##A## and ##\mathbb R^4## hence they are homemorphic.

Does it make sense ? Thanks.

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