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.