1. Since \\(\alpha_\mathrm{Arrow}(a)=d\\), it must also be the case that \\(\alpha_\mathrm{Vertex}(1)=4\\) and \\(\alpha_\mathrm{Vertex}(2)=5\\). Since \\(b:2\to3\\), \\(\alpha_\mathrm{Arrow}(b)\\) must start at \\(5\\), the only such arrow in \\(\mathcal{H}\\) is \\(e\\). So, \\(\alpha_\mathrm{Arrow}(b)=e\\) and \\(\alpha_\mathrm{Vertex}(3)=5\\).

2. The first line in \\(G(Arrow)\\) goes to the middle line of \\(H(Arrow)\\), the second line of \\(G(Arrow)\\) goes to the last line of \\(H(Arrow)\\). The first line of \\(G(Vertex)\\) goes to the first line of \\(H(Vertex)\\), the last two lines of \\(G(Vertex)\\) go to the last line of \\(H(Vertex)\\).

3. \\(s\circ\alpha_\mathrm{Arrow}(a)=s(d)=4=\alpha_\mathrm{Vertex}(1)=\alpha_\mathrm{Vertex}\circ s(a)\\). \\(s\circ\alpha_\mathrm{Arrow}(b)=s(e)=5=\alpha_\mathrm{Vertex}(2)=\alpha_\mathrm{Vertex}\circ s(b)\\). \\(t\circ\alpha_\mathrm{Arrow}(a)=t(d)=5=\alpha_\mathrm{Vertex}(2)=\alpha_\mathrm{Vertex}\circ t(a)\\). \\(t\circ\alpha_\mathrm{Arrow}(b)=t(e)=5=\alpha_\mathrm{Vertex}(3)=\alpha_\mathrm{Vertex}\circ t(b)\\). So, the matches are natural.