- #1
cianfa72
- 2,382
- 254
- 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.
Last edited: