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](https://wildtopology.wordpress.com/2014/06/28/the-griffiths-twin-cone/), $Hom(\pi\_1(\mathbb G),\mathbb Z)=0$. Is there any example of two contractible spaces whose wedge sum's fundamental group would allow a surjection onto $\mathbb Z$?



Example similar to the Griffiths twin cone but with fundamental group that allows 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.


