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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeApr 21st 2016

I have added to coequalizer basic statements about its relation to pushouts.

In the course of this I brought the whole entry into better shape.

• CommentRowNumber2.
• CommentAuthorzskoda
• CommentTimeMay 21st 2017

Something is wrong with the terminology in the idea section.

the projection function $p \colon Y \longrightarrow Y/_\sim$ satisfies

$p \circ f = p \circ g$

and in fact $p$ is universal with this property, hence it “co-equalizes” $f$ and $g$.

In the standard terminology, one says that $p$ coequalizes a parallel pair $f,g$ if $p\circ f = p\circ g$, period. No universality. (Co)equalizing is the same as making a (co)cone here, not the same as being a (co)equalizer/universal (co)cone !

• CommentRowNumber3.
• CommentAuthorTodd_Trimble
• CommentTimeMay 21st 2017
• (edited May 21st 2017)

I agree. It should read to say, “$p$ is the coequalizer of the maps $f, g$”. Edit: I made an adjustment there, and also changed the word “projection” to “quotient” since projection is given the specific meaning having to do with products.

• CommentRowNumber4.
• CommentAuthorDmitri Pavlov
• CommentTimeMar 27th 2021

Coequalizers were defined in the paper

for any finite collection of parallel morphisms. The paper refers to them as right equalizers, whereas equalizers are referred to as left equalizers.

• CommentRowNumber5.
• CommentAuthorUrs
• CommentTimeSep 4th 2021

• CommentRowNumber6.
• CommentAuthorUrs
• CommentTimeNov 8th 2021

• CommentRowNumber7.
• CommentAuthorUrs
• CommentTimeNov 8th 2021

added (here) statement of two little lemmas relating coequalizers to kernel pairs