Vtyjkyuk

The problem asks me to show that the fundamental group of the Klein bottle is generated by "latitudinal" loops $a$ and "longitudinal" loops $b$ where $a$ and $b$ obey the relation $aba^-1 = b^-1$.

The problem is that I don't understand what I'm supposed to do, since the problem is stated a bit imprecisely. I haven't covered the Seifert-Van Kampen theorem either.

Draw a short cartoon that shows the loop $aba^-1$ homotoping into $b^-1$, in the squares-with-edges-identified model.

For clarity, let the loop cross the corners one at a time, and be sure to indicate the direction of every visible segment of the loop in each frame.

(Writing down the homotopy symbolically, with coordinates, would probably be overkill).

an other approach using covering space:

Let us observe the covering space $mathbbR^2$ of $K^2$ when the covering map is the quotient map $mathbbR^2 / (x,y) sim (x+1,y) , (x,y)sim(-x,y+1) mapsto K^2 $ by $p((x,y)) =[(x,y)] $. (The cover,might be thought as a paving of $mathbbR^2$ with $K^2$'s which are "mirrored" with respect to x's axis).

Now for any loop $gamma :(I,partial I)rightarrow (K^2, y_0)$ ($y_0$ denotes the bottom-left vertex of the polygonal representation of $K^2 )$. Let us denot $tildegamma $ a lift of $gamma$ to $mathbbR^2$ such that $tildegamma(0),tildegamma(1) in p^-1(y_0)$ because $gamma$ is a loop. $tildegamma $ is homotopic to concatenation of $tildea_i_iin [l_1] , tildeb_j_jin [l_1]$ when $tildea_i , tildeb_j$ are lifts of the loops $a,b$ to points on the grid $mathbbR^2 / mathbbZ^2 = p^-1(y_0)$ . This is the part I'm not going to fully prove, but here is an intuitive sketch: (the red line is $tildegamma$ which in $mathbbR^2$ might be continuously stretched to the grid).

Now using the fact that $p_* : pi_1(mathbbR^2, x_0) rightarrow pi_1(K^2 , y_0)$ the cover map induced homomorphism is one-to-one. We get:

$$ p_*([tildea_1 * ... * tildea_l_1 * tildeb_1 * ... * tildeb_l_2 ]) = p_*([tildegamma]) Rightarrow [a]^l_1 * [b]^l_2 = [gamma] $$

Hint: you can obtain the Klein bottle from the unit square identifying its edges accordingly.

The blue one is $a$ and the red one is $b$.