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Hi, All. I am reading through McLarty’s axiomatization of category of categories, and here is a theorem that I cannot understand. It might be my silly or have forgotten some basics, but it has annoyed me for several hours.
The paper “Axiomatizing category of categories” is here
On page 1245, Theorem 5 states:
Let $f$ and $g$ be composable arrows of any category $A$. If $g$ is an identity arrow then $g\circ_A f = f$
And the proof says:
Theorem 4 plus chasing through the pushout for $g \circ_A f$.
I think to be able to use Theorem 4, we need the unique arrow induced by the pair $f,g$ to factorize through the one induced by $id_2$ and $0 \circ !_2$. if such factorization does not exist, I do not see any chance of applying theorem 4.
I think there is some condition that pushout arrow induced by of $f \circ a$ and $g \circ b$ factors though the pushout arrow induced by $a$ and $b$, and the case described in Theorem 5 of the paper satisfies the condition. If there is a condition, what is it? (I agree with the intuition that if two pairs of things have the same shape, then their pushout should have a “same fragment”, my trouble is to figure out how it formally happens.)
Thanks for any hint and attempt to help!
May I suggest editing your post to use markdown formatting for the long url? Like this [link text](http://reallyoverlylongurl.com)
Edited! Thanks for the suggestion!
I think one place to start is with the universal case, namely taking the category $A$ to be $\mathbf{3}$.
I think there are only 6 maps $2\to 3$, so I think the case $A$ is $3$ can be shown using case-by-case argument.
It seems to me the claim is:
If Theorem 3 holds for $A$ is $3$, then it for all $A$’s. I believe this is true if: 1. $3$ is indeed universal, in the following sense. 2.Theorem 5 holds for 3.
But it is not clear to me that the case for $A$ is $3$ is a universal case. Expand the “sense of universal”, it means:
For every composable pair $f:2\to A,g:2\to A$ such that $g$ factors through $1$, there is a pair of composable arrows $f_0:2\to 3,g_0:2\to 3$ such that $g0$ factors through $1$, and exists arrow $t_0:3\to A$ such that $t_0\circ f_0 = f$ and $t_0\circ g_0 = g$. I am struggling on proving this.
Thanks for correcting me if my sense of “universal” is wrong here!
Update: I just realised that I have not used the fact that $3$ retracts to $2$, I will retry tomorrow.
I have convinced myself that $3$ is indeed universal. Thank you very much for the useful hint!
Excellent, glad it helped!
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