# Example similar to the Griffiths twin cone but with fundamental group that allows surjection onto $\mathbb Z$

The Griffiths twin cone is an example of a wedge sum of two contractible spaces being non-contractible. Namely, it is the wedge sum $\mathbb G=C\mathbb H\vee_p C\mathbb H$ of two coni over the Hawaiian earring by the bad point $p\in\mathbb H$.

However, as stated at the bottom of this post, $Hom(\pi_1(\mathbb G),\mathbb Z)=0$. Is there any example of two contractible spaces whose wedge sum's fundamental group would allow a surjection onto $\mathbb Z$?

No such homomorphism is possible. The prototypical nature of the Griffiths twin cone guarantees this.

Let $$X,Y$$ be contractible with basepoints $$x,y$$ respectively. We only need to assume $$\{x\}$$ and $$\{y\}$$ are closed. Let $$X\vee Y$$ be the wedge with basepoint $$\ast$$.

Suppose $$h:\pi_1(X\vee Y)\to \mathbb{Z}$$ is non-trivial. Pick a loop $$\alpha$$ in $$X\vee Y$$ based at $$\ast$$ such that $$h([\alpha])\neq 0$$. Consider the set $$\mathscr{L}$$ of components of $$[0,1]\backslash \alpha^{-1}(\ast)$$ with the natural ordering inherited from $$[0,1]$$.

Let $$\mathscr{L}_X$$ be the set of components $$(a,b)$$ of $$[0,1]\backslash \alpha^{-1}(\ast)$$ such that $$\alpha([a,b])\subseteq X$$.

Let $$\mathscr{L}_Y$$ be the set of components $$(c,d)$$ of $$[0,1]\backslash \alpha^{-1}(\ast)$$ such that $$\alpha([c,d])\subseteq Y$$.

Since $$X$$ and $$Y$$ are both simply connected, in order to have $$[\alpha]\neq 1$$, both $$\mathscr{L}_X$$ and $$\mathscr{L}_Y$$ must be infinite. Choose enumerations $$\mathscr{L}_X=\{(a_1,b_1),(a_2,b_2),...\}$$ and $$\mathscr{L}_X=\{(c_1,d_1),(c_2,d_2),...\}$$. Let $$\mathbb{H}$$ be the Hawaiian earring and $$\ell_n$$ be the standard loop traversing the n-th circle. Let $$\mathbb{H}_{odd}$$ and $$\mathbb{H}_{even}$$ be the sub-Hawaiian earrings consisting of the odd and even indexed circles respectively.

Construct a map $$f:\mathbb{H}\to X\vee Y$$ such that $$f \circ \ell_{2n-1}= \alpha|_{[a_n,b_n]}$$ and $$f\circ \ell_{2n}= \alpha|_{[c_n,d_n]}$$ after reparameterization. Using the enumerations and ordering of $$\mathscr{L}$$, it is easy to build a loop $$\beta:[0,1]\to \mathbb{H}$$ such that $$f\circ\beta$$ is a reparmeterization of $$\alpha$$.

Since $$\mathbb{H}=\mathbb{H}_{odd}\vee \mathbb{H}_{even}$$, $$f(\mathbb{H}_{odd})\subseteq X$$, and $$f(\mathbb{H}_{even})\subseteq Y$$ where $$X$$ and $$Y$$ are contractible, $$f$$ extends to a map $$g:\mathbb{G}=C\mathbb{H}_{odd}\vee C\mathbb{H}_{even}\to X\vee Y$$ (where $$C$$ is the cone and $$\mathbb{G}$$ is the Griffiths twin cone from the question). Recalling the fact $$Hom(\pi_1(\mathbb{G}),\mathbb{Z})=0$$ from the question, we see that the composition $$h\circ g_{\#}:\pi_1(\mathbb{G})\to\mathbb{Z}$$ is trivial; However, if $$i:\mathbb{H}\to\mathbb{G}$$ is inclusion, then $$h\circ g_{\#}(i_{\#}([\beta]))=h([\alpha])\neq 0$$; a contradiction.