On 'proving' that $\langle x,y_1,\ldots,y_n | [x, y_1 \cdots y_n] = 1 \rangle \simeq \mathbb Z^2$.

Question

As the title says, I want to prove that $G:= \langle x,y_1,\ldots,y_n \mid [x, y_1 \cdots y_n] = 1 \rangle \simeq \mathbb Z^2$ algebraically.

Edit: well, this is embarrassing. That is not what I want, because it is not true (and that is why I asked this in the first place, it seemed quite suspicious). So:

Where have I messed up in my proof?

Edit': I have found the error in what follows, it is now added as an answer.

Consider a regular $2(n+1)$-gon $P$ whose edges are labeled $x,y_1,\ldots,y_n,x^{-1},y_n^{-1},\ldots,y_1^{-1}$, say, clockwise. The quotient $X$ of $P$ that identifies the edges of $\partial P$ according to this labeling gives a torus: we can "fold" $P$ so that each edge $y_i$ matches the edge labeled with its inverse. This produces a tube whose boundary corresponds to the edges labeled $x$ and $x^{-1}$ respectively, and the orientation is coherent with the torus (i.e. we do not get a Klein bottle). Hence $\pi_1(X) \simeq \mathbb Z^2$. On the other hand, a straightforward analysis via van Kampen with $U = X \setminus \{[p]\}, p \not \in \partial P$ and and $V$ (the projection of) a disk around $p$ yields $ \pi_1(X) \simeq G$.

Answer 1

Doesn’t $x\mapsto 1$ give an epinorphism from $ G$ to the free group on $n$ generators $y_1,\ldots,y_n$?

Answer 2

For completion's sake: the space $X$ is indeed a torus, the mistake was thinking that puncturing $X$ would retract to a wedge of $n+1$ copies of $S^1$ hence yielding the claimed presentation.

That is not the case; we have $U \simeq S^1 \vee S^1$ which agrees with the fact that $\pi_1(X) \simeq \mathbb Z^2$.