# Fundamental group of the wedge sum of two spaces

Let $X,Y$ be two path-connected topological spaces and $\langle A\mid R\rangle,\langle B\mid S\rangle$ respectively presentations for their fundamental groups. I think that a presentation for the fundamental group of the wedge sum $X\vee_{x_{0}} Y$ is $\langle A\sqcup B\mid R,S\rangle$.

All one should prove is that if $f$ is a loop in $X$ and $g$ a loop in $Y$ (both with base-point $x_{0}$) such that $f\cdot g\simeq g\cdot f$ in $X\sqcup Y$, then at least one of the two loops is homotopic to $x_{0}$ in his own space. How can I prove this?

I'd tried taking a homotopy $H$ between $f\cdot g$ and $g\cdot f$, and analize the possible preimages by $H$ of $x_{0}$. I saw that these preimages must contain curves of this form or similar: and this could give the needed homotopy.

I'm not sure that this reasoning is right and how to complete. Can anyone help me please? Thank you very much.

• I doubt this is true as stated: I think you'll need some sort of local regularity condition at the basepoints. – Chris Eagle Mar 4 '13 at 22:04
• You need some mild hypotheses on what a neighborhood of $x_0$ looks like, but you can prove a version of this using Seifert-van Kampen (en.wikipedia.org/wiki/Seifert%E2%80%93van_Kampen_theorem). – Qiaochu Yuan Mar 4 '13 at 22:07

The question is whether $\pi_1 : \mathsf{Top}_* \to \mathsf{Grp}$ preserves coproducts. When $x_0$ has a connected weakly contractible open neighborhood in both spaces which are connected, then it is true by the Seifert van Kampen theorem. In general it is wrong, take $X=Y=$ Hawaiian earring.
• In this example clearly the hypothesis mentioned don't hold, but why this contradicts my affirmation? The $\pi_{1}$ of the Hawaiian earring, is the free group generated by an infinite countable set or am I wrong? – Her Mar 5 '13 at 1:20
• @Her: no, $\pi_1$ of the Hawaiian earring is much more complicated. The Hawaiian earring is not a wedge sum of countably many circles; it inherits the subspace topology from $\mathbb{R}^2$ and in particular you can write down some "infinite products" of the obvious loops. – Qiaochu Yuan Mar 5 '13 at 5:30
• I personally wanted a proof to go with this statement, so here is a paper with some useful information. math.byu.edu/~conner/research/he_main.pdf Page 14-15 has clear example of $\pi_1$ not preserving coproducts – Jack Davies Nov 5 '14 at 7:27