Why is a vertex with deg⁻(v) = 0 (vertex of a directed graph with no exiting edge) called a sink?

After looking at the Example 3.2.8. (atomic graph), I got what sink means, but I didn't understand why it's called a sink. This is a term from graph theory, but I suggest adding it to the Glossary of category theory, because it is used in the definition of the operation of pasting. It seems that sink isn't just another name for codomain (target).--SilverMatsu (talk) 02:57, 12 October 2026 (UTC)

Think of a faucet (source) and a drain (sink)? —Antonissimo (talk) 04:23, 12 October 2026 (UTC)

"Sink" has many meanings describing something where things disappear. Examples (per Wiktionary:sink): sink for water, heat sink, resource sink. I've used the phrase "time sink" to any describe a time consuming activity that produces little or no tangible benefit. In any case, definitions 10 & 11 cover the graph theoretical meanings, and I think the analogy is pretty clear. --RDBury (talk) 16:09, 12 October 2026 (UTC)

I see. Thank you.--SilverMatsu (talk) 02:36, 14 October 2026 (UTC)

The article Pasting diagram is not comprehensible in its current state, since it uses the notion of "pasting diagram" throughout without ever defining it. It contains a definition of "2-pasting diagram", after the term "(pasting) diagram" has already been used 10 times, but this definition requires an understanding of the concept of "admitting a pasting scheme presentation", which also lacks a definition.

The term sink as used by Johnson & Yau is ancillary to their definition of an "anchored" (interior) face of a planar connected graph. It is not a category-theoretical term. In standard graph-theoretical parlance, the term is used in directed graphs, but Johnson & Yau use it to induce a direction on paths in an undirected graph, which is not standard at all.  ​‑‑Lambiam 07:40, 12 October 2026 (UTC)

By the way, where do you find Example 3.2.8.? —Antonissimo (talk) 17:41, 12 October 2026 (UTC)

In Johnson, Niles; Yau, Donald (2021). 2-Dimensional Categories. p. 74. arXiv:2002.06055, referenced below as well as in Pasting diagram.  ​‑‑Lambiam 21:52, 12 October 2026 (UTC)

Inverting the Nicolosi map projection

Among the imperfect ways to map locations on a spherical globe onto a flat sheet the Nicolosi projection has historical interest and features a balanced compromise of distortions. Wikipedia's article gives the formula for transforming each (longitude:latitude) globe location to a rectangular (x:y) map coordinate. The formula is in a direct form that is readily calculable with only simple correction needed at the equator and around the edge of the map. I assume that a direct-form inverse of the Nicolosi 2-D transform can exist. (Using Newton's method in 2-D to refine initial guesses is an option but this is a costly iterative approach poorly suited to efficient computer graphics.) The chatbot CoPilot supported my assumption but wimped out on providing the inverse transform. Is there a human mathematician here who will provide the formula?~2026-55298-71 (talk) 14:25, 15 October 2026 (UTC)

Looking with half an eye at the equations, I spot,

${\displaystyle {\begin{aligned}c&={\frac {2\varphi }{\pi }}\\d&={\frac {1-c^{2}}{\sin \varphi -c}}.\end{aligned}}}$

The variable ${\displaystyle c}$ is used solely in the right-hand side of the equation defining ${\displaystyle d.}$ We can therefore eliminate ${\displaystyle c}$ from the equation set and combine these two equations into

${\displaystyle d=f(\varphi )}$

where

${\displaystyle f(\varphi )={\frac {1-{\frac {2\varphi }{\pi }}^{2}}{\sin \varphi -{\frac {2\varphi }{\pi }}}}.}$

It seems to me that inverting the equations in closed form is only possible if this function ${\displaystyle f}$ can be inverted in closed form. However, this can't be done with the arsenal of standard functions.

When using Newton–Raphson, this can be made faster by using the earlier solution for a neighbouring pixel to provide a good approximation (in most cases) of the initial values for a subsequent pixel. You could even extrapolate the values of the last two earlier pixels in a row to cheaply get even better initial approximations.  ​‑‑Lambiam 06:03, 16 October 2026 (UTC)

I suspect you found inverting the sine term in the denominator daunting. Considering that it is applied only to latitudes within the range +/- 1.5708 radian and that we don't require from Nicolusi the high precisions of navigation or surveying since the map transform is mainly for subjective effect, it may help to replace the sine function with its truncated Taylor series. In BASIC the simplification is

 sin(x)  <----  ( x - x^3 / 6 ) 

I tested this at the point in the formula that you spotted. The simplification theoretically introduces less than 6.8° errors near the north and south poles which I think is easily tolerable. ~2026-55298-71 (talk) 13:39, 17 October 2026 (UTC)

It is not clear what you'd use the inverse formulas for and you did not make clear that a rough-and-ready approximation would already be satisfactory. But even so, obtaining one will be a highly laborious effort, even if you approximate the equations by polynomial ones that can be solved algebraically. I strongly doubt that cubic approximations for the trigonometric functions will help in finding usable formulas.

Considering the function given by

${\displaystyle s(x)=x-{\frac {x^{3}}{6}},}$

it never reaches ${\displaystyle 1}$ for ${\displaystyle x>0}$ but attains a maximal value of ${\displaystyle {\tfrac {2}{3}}{\sqrt {2}}\approx 0.94281}$ for ${\displaystyle x={\sqrt {2}},}$ corresponding to ${\displaystyle 81^{\circ }.}$ It is not usable as such near the poles. You could use instead

${\displaystyle s(x)={\frac {3\pi ^{2}x-4x^{3}}{\pi ^{3}}},}$

for which ${\displaystyle s^{{-}1}(1)={\tfrac {1}{2}}\pi .}$ Then ${\displaystyle x'=s^{{-}1}(\sin x)}$ differs from ${\displaystyle x}$ by no more than ${\displaystyle 0.0285,}$ corresponding to a difference of about ${\displaystyle 1.63^{\circ }.}$ But, as stated before, I don't expect that this will help much for the larger problem.  ​‑‑Lambiam 19:45, 17 October 2026 (UTC)

It is not necessary to solve numerically in 2D, since it is simple to find the circumcentre of the poles and the given point. The intersection of equator and the arc through these points gives λ immediately. catslash (talk) 22:34, 17 October 2026 (UTC)

I'm sorry, but I cannot decipher this. The Nicolosi globular projection provides a mapping ${\displaystyle (\varphi ,\lambda )\mapsto (x,y).}$ The inverse would be a mapping ${\displaystyle (x,y)\mapsto (\varphi ,\lambda ).}$ What is the meaning of solving this "in 2D"? Also, which of the four points referred to in your reply are "these points", and is "the arc through these points" meant to be a circular arc?  ​‑‑Lambiam 10:46, 18 October 2026 (UTC)

I now see what you mean: all points of a given longitude lie on a circle (for ${\displaystyle \lambda =\lambda _{0}}$ a straight line through the poles) that intersects the equator at that longitude. That gives us the part ${\displaystyle (x,y)\mapsto \lambda .}$ Since the points of a given latitude also lie on a circle or straight line, similarly determined by its intersections, but now with the central vertical and the circular boundaries, this should similarly lead to ${\displaystyle (x,y)\mapsto \varphi ,}$ although this is less simple. These two parts of the inversion can be computed independently.  ​‑‑Lambiam 20:13, 18 October 2026 (UTC)

Might it help to express the Nicolosi transform as a 2D Fourier transform? The coefficients of a Fourier transform also serve in its inverse transform with a change of a constant factor. ~2026-55298-71 (talk) 14:35, 19 October 2026 (UTC)