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1. • CommentRowNumber1.
   • CommentAuthorUrs
   • CommentTimeNov 18th 2010
   • PermaLink
   Author: Urs
   Format: MarkdownItexedited [[retract]] a little

   edited retract a little

2. • CommentRowNumber2.
   • CommentAuthorHarry Gindi
   • CommentTimeNov 19th 2010
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   Author: Harry Gindi
   Format: MarkdownItexI think we should emphasize that splittings and retractions are subtly different. One is a "left inverse to an inclusion" and the other is a "right inverse to a projection" (I may have gotten my "handedness" screwed up, but I'm sure you get the idea).

   I think we should emphasize that splittings and retractions are subtly different. One is a “left inverse to an inclusion” and the other is a “right inverse to a projection” (I may have gotten my “handedness” screwed up, but I’m sure you get the idea).

3. • CommentRowNumber3.
   • CommentAuthorDavidRoberts
   • CommentTimeNov 19th 2010
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   Author: DavidRoberts
   Format: MarkdownItexGo ahead, then...

   Go ahead, then…

4. • CommentRowNumber4.
   • CommentAuthorzskoda
   • CommentTimeJan 26th 2011
   • (edited Jan 26th 2011)
   • PermaLink
   Author: zskoda
   Format: MarkdownItexIn homotopy theory there is also the terminology space $X$ is dominated by space $Y$ (or by a class of spaces) if $X$ is a retract of $Y$ in the homotopy category. I do not know how to name an entry (domination is a used term but more rarely and more artifically than the phrase $X$ is dominated by $Y$). There is also "shape dominated by" and alike variants. Edit: maybe one should write the section under retract and just put redirects like "dominated by", though retract is general categorical page and we are here in more specific homotopy context.

   In homotopy theory there is also the terminology space $X$ is dominated by space $Y$ (or by a class of spaces) if $X$ is a retract of $Y$ in the homotopy category. I do not know how to name an entry (domination is a used term but more rarely and more artifically than the phrase $X$ is dominated by $Y$). There is also “shape dominated by” and alike variants.

   Edit: maybe one should write the section under retract and just put redirects like “dominated by”, though retract is general categorical page and we are here in more specific homotopy context.

5. • CommentRowNumber5.
   • CommentAuthorTobyBartels
   • CommentTimeJan 26th 2011
   • (edited Jan 26th 2011)
   • PermaLink
   Author: TobyBartels
   Format: MarkdownItexIf you want a page title that fits with the naming conventions, maybe [[dominated space]]? That makes it sound like a kind of space (analogous to, say, [[connected space]]), which it is not. But English grammar is more flexible than that interpretation. But if you put it on [[retract]], then you can add any phrases as redirects, nouns or otherwise, as long as they don't seem to mean anything else. (See the problems with some [adjectival redirects](http://www.math.ntnu.no/~stacey/Mathforge/nForum/comments.php?DiscussionID=2412&page=1).)

   If you want a page title that fits with the naming conventions, maybe dominated space? That makes it sound like a kind of space (analogous to, say, connected space), which it is not. But English grammar is more flexible than that interpretation.

   But if you put it on retract, then you can add any phrases as redirects, nouns or otherwise, as long as they don’t seem to mean anything else. (See the problems with some adjectival redirects.)

6. • CommentRowNumber6.
   • CommentAuthorgejza.jenca
   • CommentTimeMar 29th 2016
   • (edited Mar 29th 2016)
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   Author: gejza.jenca
   Format: MarkdownItexYesterday I stumbled upon the following fact: if the underlying category $\mathcal{C}$ has coproducts, then retractions can be considered as algebras for a monad on the arrrow category $\mathcal{C}^{\,2}$. Let $f:A\to B$ be an object of $\mathcal{C}^{\,2}$. Define $T(f):A\oplus B\to B$ to be the arrow determined by the cocone $f,id_B$ -- this is the free retraction generated by $f$. The $\eta$ and $\mu$ are then the only possible thing. This is a monad. It turns out that, given $f$, there exists an algebra (commutative square) $T(f)\to f$ if and only if $f$ is a retraction. Moreover, the algebra is determined by a choice of section $i:B\to A$. I would like to add this information to nlab, but I am not sure where it belongs. There are at least three different pages dealing with retracts: [[retract]], [[retraction]], [[section]].

   Yesterday I stumbled upon the following fact: if the underlying category $\mathcal{C}$ has coproducts, then retractions can be considered as algebras for a monad on the arrrow category $\mathcal{C}^{\,2}$.

   Let $f:A\to B$ be an object of $\mathcal{C}^{\,2}$. Define $T(f):A\oplus B\to B$ to be the arrow determined by the cocone $f,id_B$ – this is the free retraction generated by $f$. The $\eta$ and $\mu$ are then the only possible thing. This is a monad. It turns out that, given $f$, there exists an algebra (commutative square) $T(f)\to f$ if and only if $f$ is a retraction. Moreover, the algebra is determined by a choice of section $i:B\to A$.

   I would like to add this information to nlab, but I am not sure where it belongs. There are at least three different pages dealing with retracts: retract, retraction, section.

7. • CommentRowNumber7.
   • CommentAuthorUrs
   • CommentTimeMar 29th 2016
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   Author: Urs
   Format: MarkdownItexThanks. The keywords _[[retract]]_ and _[[retraction]]_ should better point to the same entry. I have merged them now and made both keywords point to the same entry. That entry would be the canonical place to add material.

   Thanks. The keywords retract and retraction should better point to the same entry. I have merged them now and made both keywords point to the same entry. That entry would be the canonical place to add material.

8. • CommentRowNumber8.
   • CommentAuthorTodd_Trimble
   • CommentTimeMar 29th 2016
   • (edited Mar 29th 2016)
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   Author: Todd_Trimble
   Format: MarkdownItexI don't think that's quite the canonical place. Better IMO would be [[co-slice category]] or [[undercategory]]; such an observation likely exists in dual form on the Lab at [[slice category]], since people use it all the time. In other words, $A \downarrow \mathcal{C}$ is the category of algebras of the monad $A + -$, just as a slice category $\mathcal{C} \downarrow A$ is the category of coalgebras for the comonad $A \times -$.

   I don’t think that’s quite the canonical place. Better IMO would be co-slice category or undercategory; such an observation likely exists in dual form on the Lab at slice category, since people use it all the time.

   In other words, $A \downarrow \mathcal{C}$ is the category of algebras of the monad $A + -$, just as a slice category $\mathcal{C} \downarrow A$ is the category of coalgebras for the comonad $A \times -$.

9. • CommentRowNumber9.
   • CommentAuthorUrs
   • CommentTimeJun 28th 2018
   • PermaLink
   Author: Urs
   Format: MarkdownItexmade explicit the lemma ([here](https://ncatlab.org/nlab/show/retract#LeftInverseWithLeftInverseIsLeftInverse)) that a left inverse with a left inverse is an inverse. <a href="https://ncatlab.org/nlab/revision/diff/retract/22">diff</a>, <a href="https://ncatlab.org/nlab/revision/retract/22">v22</a>, <a href="https://ncatlab.org/nlab/show/retract">current</a>

   made explicit the lemma (here) that a left inverse with a left inverse is an inverse.

   diff, v22, current

10. • CommentRowNumber10.
    • CommentAuthorsegfau1t
    • CommentTimeMar 28th 2024
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    Author: segfau1t
    Format: MarkdownItexI'm just poking around trying to learn, but this page seems to imply that composing section and retraction "backwards," $s \circ r$ is idempotent. This seems simple to show, could/should it be added to properties?

    I’m just poking around trying to learn, but this page seems to imply that composing section and retraction “backwards,” $s \circ r$ is idempotent. This seems simple to show, could/should it be added to properties?

11. • CommentRowNumber11.
    • CommentAuthorJ-B Vienney
    • CommentTimeMar 28th 2024
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    Author: J-B Vienney
    Format: MarkdownItexThis is absolutely true. This fact is hidden in the definition under the form "The entire situation is said to be a splitting of the idempotent $i \circ r$."

    This is absolutely true. This fact is hidden in the definition under the form “The entire situation is said to be a splitting of the idempotent $i \circ r$.”