# Start a new discussion

## Not signed in

Want to take part in these discussions? Sign in if you have an account, or apply for one below

## Discussion Tag Cloud

Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.

• 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.