Howdy, Stranger!
It looks like you're new here. If you want to get involved, click one of these buttons!
Categories
- All Categories 2.2K
- Programming with Categories Course 21
- Exercises - Programming with Categories Course 15
- Applied Category Theory Course 341
- Applied Category Theory Seminar 4
- Exercises - Applied Category Theory Course 149
- Discussion Groups 50
- How to Use MathJax 15
- Chat 487
- Azimuth Code Project 108
- News and Information 147
- Azimuth Blog 149
- Azimuth Forum 29
- Azimuth Project 189
- - Strategy 108
- - Conventions and Policies 21
- - Questions 43
- Azimuth Wiki 711
- - Latest Changes 701
- - - Action 14
- - - Biodiversity 8
- - - Books 2
- - - Carbon 9
- - - Computational methods 38
- - - Climate 53
- - - Earth science 23
- - - Ecology 43
- - - Energy 29
- - - Experiments 30
- - - Geoengineering 0
- - - Mathematical methods 69
- - - Meta 9
- - - Methodology 16
- - - Natural resources 7
- - - Oceans 4
- - - Organizations 34
- - - People 6
- - - Publishing 4
- - - Reports 3
- - - Software 21
- - - Statistical methods 2
- - - Sustainability 4
- - - Things to do 2
- - - Visualisation 1
- General 41
Options
Exercise 62 - Chapter 3
We claim that there is exactly one graph homomorphism \( \alpha : G \rightarrow H \) such that \( \alpha_{Arrow}(a) = d \).
- What is the other value of \( \alpha_{Arrow}\), and what are the three values of \( \alpha_{Vertex} \)?
- Draw \( \alpha_{Arrow} \) as two lines connecting the cells in the ID column of \(G(Arrow)\) to those in the ID column of \(H(Arrow)\). Similarly, draw \( \alpha_{Vertex} \) as connecting lines.
- Check the source column and target column and make sure that the matches are natural, i.e. that “alpha-then-source equals source-then-alpha” and similarly for “target”.
$$ \begin{matrix} G := & & \begin{array}{c|cc} Arrow & source & target \\ \hline a & 1 & 2 \\ b & 2 & 3 \end{array} & & \begin{array}{c|} Vertex \\ \hline 1 \\ 2 \\ 3 \end{array} \\ H := & \begin{array}{c|cc} Arrow & source & target \\ \hline c & 4 & 5 \\ d & 4 & 5 \\ e & 5 & 5 \end{array} & & \begin{array}{c|} Vertex \\ \hline 4 \\ 5 \end{array} & \end{matrix} $$
Comments
The graph instances are offset so you can more easily complete the exercise.
The graph instances are offset so you can more easily complete the exercise.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 \(\alpha_\mathrm{Vertex}(3)=5\).
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 \(H(Vertex)\).
\(s\circ\alpha_\mathrm{Arrow}(a)=s(d)=4=\alpha_\mathrm{Vertex}(1)=\alpha_\mathrm{Vertex}\circ s(a)\). \(s\circ\alpha_\mathrm{Arrow}(b)=s(e)=5=\alpha_\mathrm{Vertex}(2)=\alpha_\mathrm{Vertex}\circ s(b)\). \(t\circ\alpha_\mathrm{Arrow}(a)=t(d)=5=\alpha_\mathrm{Vertex}(2)=\alpha_\mathrm{Vertex}\circ t(a)\). \(t\circ\alpha_\mathrm{Arrow}(b)=t(e)=5=\alpha_\mathrm{Vertex}(3)=\alpha_\mathrm{Vertex}\circ t(b)\). So, the matches are natural.
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.