Timeline for On internal functions and arrows in a Topos
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Mar 25, 2013 at 9:51 | comment | added | Wouter Stekelenburg | To see that $\pi_0: L\to X$ is a monomorphism, consider its kernel pair $l,r:Z\to L\times L$. Now $\pi_0\circ l =\pi_0\circ r$ by definition, and $\pi_1\circ l = \pi_1\circ r$ by uv), because the morphism $(\pi_0\circ l,\pi_1\circ l,\pi_1\circ r): Z\to X\times Y\times Y$ represents the subobject $\{ (x,y,y')| \lambda(x,y)\land\lambda(x,y') \}$. Because $\pi_i\circ l = \pi_i\circ r$, $l=r$. To conclude: every parallel pair $f,g:W\to L$ such that $\pi_0\circ f= \pi_0\circ g$ factors uniquely through $l,r$. Therefore every such pair satisfies $f=g$. | |
| Jun 27, 2012 at 17:54 | comment | added | Wouter Stekelenburg | Left exact functors preserve posets, and $\Omega_F$ is an internal poset in $\mathcal F$. The lattice structures of $\mathcal E(X\times Y,f_*\Omega)$ and $\mathcal F(f^*X\times f^*Y,\Omega)$ coincide because they are the result of applying naturally isomorphic left exact functors -- $\mathcal E(X\times Y,f_*-)$ and $\mathcal F(f^*X\times f^*Y,−)$ -- to the same poset. This equivalence in natural in $X$ and $Y$. Now joins have a universal property relative in terms of the order, and therefore must commute with the isomorphism. | |
| Jun 27, 2012 at 11:41 | history | edited | Wouter Stekelenburg | CC BY-SA 3.0 |
edited body
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| Jun 26, 2012 at 21:35 | history | edited | Wouter Stekelenburg | CC BY-SA 3.0 |
corrected symbol
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| Jun 25, 2012 at 21:28 | history | edited | Wouter Stekelenburg | CC BY-SA 3.0 |
added 172 characters in body
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| Jun 25, 2012 at 8:09 | history | answered | Wouter Stekelenburg | CC BY-SA 3.0 |