The document discusses a projection map pi_1 from R^2 to R that projects onto the first component. It defines the subspace A of R^2 as the union of points where x is greater than or equal to 0 and points where y is equal to 0. The projection map pi_1 restricts to an open mapping from the subspace A to R because the projection of the first component x maps from -infinity to +infinity, which is an open interval. However, the projection map is not a closed mapping, as shown by considering the image of the tangent function under projection, which is always an open, not closed, interval.

1. Let pi_1: R^2 rightarrow R be the projection map onto the first component and A the
subspace{(x, y): x greaterthanorequalto 0} U {(x, y): y = 0} of R^2
Solution
The Domain of R^2 is that both x and y are real number mapping from -inf to + inf which is
open
f(U) = projection of this function in R where projection of first component x maos from -inf to
+inf
So F(U) is a open mapping.
Hence this is an open map.
This mapping is not a closed mapping.
Take image of tan(x).. The projection will always be an open interval.
S it is not a closed map.