关于去除两点的克莱因瓶中道路α的同伦类计算的技术问询

Hey there! Great question—you’re already off to a solid start recognizing the space is homotopy equivalent to $S^1 \vee S^1 \vee S^1$. Let’s walk through both approaches you mentioned, since both are totally feasible with your current toolkit (Van Kampen’s theorem is exactly the right tool here).

Using the Klein Bottle Square Representation #

If you’re comfortable with the standard square identification for the Klein Bottle, this can be the faster method. Here’s how to go about it:

• First, recall the fundamental group generators for the Klein Bottle itself: usually denoted $a$ and $b$, corresponding to the two glued edges of the square (stick to the direction conventions from your reference—e.g., $a$ might be the horizontal edge glued left-right with a twist, $b$ the vertical edge glued top-bottom).
• When you remove two points $Q$ and $R$, think of them as two small "holes" punched into the square. Each hole adds a new generator to the fundamental group (let’s call one of them $c$)—per Van Kampen’s theorem, gluing a neighborhood around each hole (a disk minus a point, homotopy equivalent to $S^1$) adds free generators, and since the space is homotopy equivalent to three wedged circles, we end up with a free group on 3 total generators.
• Trace your path $\alpha$ on the square and break it into segments:
  • Segments following the glued edges translate directly to $a$, $a^{-1}$, $b$, or $b^{-1}$ depending on direction.
  • Segments looping around a removed point translate to $c$, $c^{-1}$, or the inverse based on loop direction.
• Concatenate these generators in the order the path traverses them, and that’s your homotopy class in the free group $F(a,b,c)$.

Using the Intuitive Wedge of Circles #

If the square’s glued edges feel confusing, visualizing the space as three wedged circles is super intuitive:

• Think of the Klein Bottle minus two points as three circles stuck together at a single basepoint. Each circle corresponds to one generator of the fundamental group: two from the original Klein Bottle’s core loops, one from the combined effect of the two removed points.
• Trace $\alpha$ on this wedge diagram: count how many times it loops around each circle, and in which direction. For example, if $\alpha$ goes around the first circle clockwise once, the second counter-clockwise twice, then the third clockwise once, your homotopy class is $a b^{-2} c$.
• This method bypasses the square’s glueing rules entirely—you just map the path directly to the free group elements of the wedge’s fundamental group.

Which Approach is Easier? #

It totally depends on your comfort level:

• If you have the square’s generator directions memorized and can easily decompose $\alpha$ into square segments, go with that—it’s more direct from the original space’s definition.
• If glueing directions make your head spin, the wedge visualization is simpler, since you’re working with a space whose fundamental group is a free group with no relations (unlike the Klein Bottle itself, which has a relation between $a$ and $b$).

Just remember: since the space is homotopy equivalent to $S^1 \vee S^1 \vee S^1$, its fundamental group is a free group on three generators—any path’s homotopy class is just a word in those generators, and both methods will get you that word.

备注：内容来源于stack exchange，提问作者fish