# Fundamental group of Klein Bottle from $S^1 \vee S^1 \vee S^2$?

Klein Bottle $K$ is homotopy equivalent to

$S^1 \vee S^1 \vee S^2$.

Intuitively then $\pi_1(K) \cong \langle c,d \rangle$. However this is wrong, because

$\pi_1(K) \cong \langle c,d \rangle/cdc^{-1}d$.

Can anybody explain why this intuition fails? Thanks.

I can't understand why homotopy equivalence does not preserve the homotopy group? I would like to see an intuitive explanation, not based on abstract reasoning.

• To my mind, intuitive explanations are usually somewhat abstract. Of course, abstract reasonings can be quite unintuitive. – Arthur Jul 26 '17 at 9:37

Read your link again. They're saying that a subspace of $\mathbb{R}^3$ (formed by a non-injective continuous image of $K$) is homotopy equivalent to $S^1 \vee S^1 \vee S^2$, not $K$ itself.
• Not necessarily. There is an injective map $[0,1) \to S^1$ which is also surjective, hence the image has fundamental group $\mathbb{Z}$. – Dan Rust Jul 26 '17 at 9:45
• But $[0,1)$ is not a topological space. By a space I meant a topological space, sorry for confusion. – Tyrell Jul 26 '17 at 9:46
• $[0,1)$ is certainly a topological space, given by the subspace topology inherited from $\mathbb{R}$ (for instance). – Dan Rust Jul 26 '17 at 9:46
• $[0,1)$ is not homotopy equivalent to $S_1$. $[0,1)$ is contractable to a point, and $S_1$ is not. So I can't understand how this qualifies as an example. – Tyrell Jul 26 '17 at 9:56