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# Daniel Wang

• $$G_1$$: $$f = g$$ $$G_2$$: None (I think) $$G_3$$: $$f.h = g.i$$ $$G_4$$: None
Comment by Daniel Wang June 2018
• The compositions will both map to $$f . h$$.  \begin{array}{c|c} \text{morphism }f & F~f \\hline \text{id}_A' & \text{id}_A \ \text{id}_B' & \text{id}_B \ \text{id}_C' & \text{id}_C \ \text{id}_D' & \text{id}_D \ f' & f \…
Comment by Daniel Wang June 2018
• Both $$F: C \to D$$ and $$G: C \to D$$ can map the left point to the left point, and the right point to the right point. But they can differ by having $$F$$ map the one morphism in $$C$$ to the top morphism in $$D$$ and having $$G$$ map the one mor…
Comment by Daniel Wang June 2018
• I was curious how the choice of symmetric monoidal poset affects the resulting $$\mathbf{Bool}$$-enriched category. From any poset over $$\mathbf{Bool}$$, we can make it symmetric monoidal by picking an $$I$$ and a $$\otimes$$. For $$\leq$$, I will…
• @MichaelHong: What do you use to draw your diagrams? They're staggeringly beautiful.
Comment by Daniel Wang April 2018
• I think once you understand the goal (which John helpfully pointed out) it becomes much simpler. I was able to solve this with just applications of Proposition 1.81 (and reflexivity). Since the process is nearly unchanged for the case of left adjoin…
Comment by Daniel Wang April 2018
• $$f(x) = x^5$$ is order-preserving, but $$f(x) = |x|$$ is not; $$f(x) = -x$$ is metric-preserving, but $$f(x) = 1$$ is not; $$f(x) = \frac{x}{2}$$ is addition-preserving, but $$f(x) = |x|$$ is not.
Comment by Daniel Wang April 2018