Question

Boundary circle of mobius band is not retract to its

Answer 1

Step-by-step explanation:

Suppose there were a retract r:M→∂M. By definition, this means that r∘i=id∂M, where i:∂M→M is the inclusion. From functoriality, it follows that r∗∘i∗=idπ1(∂M), where f∗ denotes the induced map of fundamental groups. Thus, r∗:π1(M)→π1(∂M) is surjective. However, π1(M)≅Z≅π1(∂M) and r∗(n)=2n, which is not surjective: a contradiction. Thus, there can be no such retract.

To see that r∗(n)=2n, I think it is easiest to view the Möbius strip as a quotient of the unit square in R2, obtained by identifying the left and right sides with the opposite orientation. Intuitively, if you go around the Möbius band once you, the projection onto the boundary goes around twice (draw a picture for yourself).