Second group homotopy CW complex relative 1-skeleton

Solution 1:

As $X^1$ is a connected graph, you know $\widetilde{X^1}$ is a tree which is contractible. Thus $\pi_2(X^1)=\pi_2(\widetilde{X^1})=0$, and the long exact sequence that you mention restricts to a short exact sequence of groups: $$1\to \pi_2(X)\to\pi_2(X,X^1)\to \ker(i_*)\to 1$$ Again as $X^1$ is a graph we have that $\pi_1(X^1)$ is a free group, so ker$(i_*)$ is a subgroup of a free group, hence itself free.

Any short exact sequence of groups, where the third term is a free group, will split. Thus the above short exact sequence splits.

It remains to show that $\pi_2(X)$ is central in $\pi_2(X,X^1)$. Let $\alpha\in \pi_2(X)$ and $\beta\in\pi_2(X,X^1)$. We show $\beta\alpha\beta^{-1}=\alpha$ diagramatically: