Wikipedia:Reference desk/Archives/Mathematics/2014 September 3

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September 3

Coequalizer, followup

Via google I can't find examples, and my brain bug that prevents me from seeing things clear is not really dead yet. What do coequalizers in the arrow category of sets look like? 93.132.23.10 (talk) 12:44, 3 September 2014 (UTC)

If my question was too special or too complex or too difficult, where can I get some reading about the arrow category of sets?

77.3.171.164 (talk) 14:38, 6 September 2014 (UTC)

And any reading, or examples of, coequalizers in other categories will be welcome, too. 77.3.171.164 (talk) 14:54, 6 September 2014 (UTC)

Any two functions $f_{X}:X_{1}\rightarrow X_{2}$ and $f_{Y}:Y_{1}\rightarrow Y_{2}$ are objects in the arrow category of sets. Let $(g_{1},g_{2})$ and $(h_{1},h_{2})$ be morphisms from $f_{X}$ to $f_{Y}$, that is $g_{1},h_{1}:X_{1}\rightarrow Y_{1}$ and $g_{2},h_{2}:X_{2}\rightarrow Y_{2}$ such that $f_{Y}\circ g_{1}=g_{2}\circ f_{X}$ and $f_{Y}\circ h_{1}=h_{2}\circ f_{X}$.

Now construct the coequalizers in the category of sets of the pairs of morphisms $g_{1},h_{1}$ and $g_{2},h_{2}$ as in your previous question, i.e. let $\sim _{1}$ and $\sim _{2}$ be the smallest equivalence relations on $Y_{1}$ and $Y_{2}$ such that $g_{1}(x)\sim _{1}h_{1}(x)$ for all $x\in X_{1}$ and $g_{2}(x)\sim _{2}h_{2}(x)$ for all $x\in X_{2}$, let $Q_{1}=Y_{1}/\sim _{1}$ and $Q_{2}=Y_{2}/\sim _{2}$ and define $q_{1}:Y_{1}\rightarrow Q_{1}$ by $q_{1}(y)=[y]_{1}$ for all $y\in Y_{1}$ and $q_{2}:Y_{2}\rightarrow Q_{2}$ by $q_{2}(y)=[y]_{2}$ for all $y\in Y_{2}$ where $[y]_{1}$ and $[y]_{2}$ denote equivalence classes of $\sim _{1}$ and $\sim _{2}$.

Now we've got all that notation we can construct the coequalizer of the pair of morphisms $(g_{1},g_{2}),(h_{1},h_{2})$. The hard part is to show that for all $y,y'\in Y_{1}$, if $y\sim _{1}y'$ then $f_{Y}(y)\sim _{2}f_{Y}(y')$. I'm short of time right now so I won't prove this. It follows that we can define a function $f_{Q}:Q_{1}\rightarrow Q_{2}$ by $f_{Q}([y]_{1})=[f_{Y}(y)]_{2}$ for all $y\in Y_{1}$. Then $f_{Q}$ together with the morphism $(q_{1},q_{2})$ from $f_{Y}$ to $f_{Q}$ is the coequalizer of $(g_{1},g_{2}),(h_{1},h_{2})$. 121.99.220.36 (talk) 22:47, 6 September 2014 (UTC)

Thanks a lot. It always strikes me how simple those things turn out to be when I'm still not able to get those ideas without help. So what you do is you construct the coequalizers for source and target separately and then put them together so that this works in the arrow category. I guess it's a good exercise for me to fill in the missing proof. Thanks again. 95.112.218.246 (talk) 09:25, 7 September 2014 (UTC)

Why is this surface area approximation for a solid of revolution erroneous?

The task is rather simple: find the surface area over the unit circle centered at the origin of the paraboloid $z=x^{2}+y^{2}$, the solid of revolution obtained by revolving the parabola $z=x^{2}$ around the z-axis. Denote the unit circle as R for the double integral. The correct computation is using the double integral $S=\iint \limits _{R}{\sqrt {1+(2x)^{2}+(2y)^{2}}}dxdy=2\pi \int _{0}^{1}{\sqrt {1+4r^{2}}}rdr={\frac {\pi }{6}}(5{\sqrt {5}}-1)$ (using polar coordinates). This is also in agreement with a solid of revolution formula $S=\int _{x=0}^{x=1}2\pi xds=\int _{0}^{1}2\pi x{\sqrt {1+({\frac {dy}{dx}})^{2}}}dx=\int _{0}^{1}2\pi x{\sqrt {1+4x^{2}}}dx$ (ds is the arc length element of the paraboloid along a radial plane passing through the origin and parallel to the z-axis).

The problem is, why doesn't the following approximation work? Naively one may expect it to be the same, but it is not. Approximate the surface area using the sum of infinitely many cylindrical sections with height dz and circumference $2\pi x=2\pi {\sqrt {z}}$, whose surface area is each $2\pi {\sqrt {z}}dz$. Since z varies from 0 to 1 as x varies from 0 to 1, the erroneous notion was that $S=\int _{0}^{1}2\pi {\sqrt {z}}dz$, which is obviously wrong. My hunch is that this cylindrical approximation becomes very poor near the origin, where the paraboloid has a large horizontal component to it.--Jasper Deng (talk) 19:38, 3 September 2014 (UTC)

Instead of discarding the horizontal component ${\frac {1}{2{\sqrt {z}}}}dz$ of your paraboloid section to get a cylinder of radius ${\sqrt {z}}$ and height dz, you should "twist" the paraboloid section to get a cylinder of radius ${\sqrt {z}}$ and height ${\sqrt {1+{\frac {1}{4z}}}}dz$. Egnau (talk) 20:36, 3 September 2014 (UTC)

The approximation is not following the surface closely enough. It's analogous to approximating a diagonal with a staircase of vertical and horizontal segments. In that case the approximation can be arbitrarily close to the curve and can be used to approximate the area underneath, but if you try to use it for length it will be incorrect and it doesn't help to use more segments. (This idea was shown to me as a "proof" that √2 = 2.) In order to get an approximation to the length of a curve or the area of a surface, the approximating curve (resp. surface) has to approximate the tangent line (resp. plane) of the original, not just the position. --RDBury (talk) 11:15, 5 September 2014 (UTC)

You might be interested in Pappus's centroid theorem which deals with this sort of problem. Dmcq (talk) 13:44, 13 September 2014 (UTC)