Let pi1 R2 rightarrow R be the projection map onto the first

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.