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Stream: learning: questions

Topic: Pullbacks and coequalizers

James Deikun (Oct 08 2024 at 00:47):

I know pullbacks and coequalizers don't commute in general. My question is: is there a nice condition (sufficient, not necessarily necessary) under which the coequalizers of a "pullback of parallel pairs" form a pullback square themselves?

James Deikun (Oct 08 2024 at 00:51):

(Context: I'm trying to figure out when a certain functor defined using a coequalizer preserves pullbacks!)

James Deikun (Oct 08 2024 at 03:07):

More context:

It's the coequalizer of $TA \cdot R \; \genfrac{}{}{0pt}{0}{\xrightarrow[\hphantom{a \cdot R}]{\beta_A}}{\xrightarrow[a \cdot R]{\hphantom{\beta_A}}} A \cdot R$.

$R$ is a $T$-algebra, $\cdot$ is copower, $\beta$ is a right module over $T$ on $- \cdot R$, $a : TA \to A$ is a deconstructed $T$-algebra. It's a reflexive coequalizer, if that helps. And you can assume $T$ will always preserve pullbacks.

James Deikun (Oct 08 2024 at 03:19):

So I basically have the freedom to choose $R$ and $\beta$ to make this preserve pullbacks, but $A$ and $a$ are arbitrary. (They're the thing whose pullbacks I am trying to preserve.)