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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeJan 12th 2011

I have split off effective epimorphism in an (infinity,1)-category from effective epimorphism and polished and expanded slightly.

• CommentRowNumber2.
• CommentAuthorUrs
• CommentTimeApr 25th 2013

I have made at effective epimorphism in an (infinity,1)-category the characterization in an infinity-topos by “induces epi on connected components” more explicit.

This was in reaction to an MO question “What is the homotopy colimit of the Cech nerve as a bi-simplical set? “. However, when I was done compiling my reply, the question had been deleted, it seems.

• CommentRowNumber3.
• CommentTimeApr 25th 2013

The question seems to be on math.stackexchange still.

• CommentRowNumber4.
• CommentAuthorUrs
• CommentTimeApr 25th 2013

Ah, thanks!

You give an excellent reply there. I have just added a comment now on where to find this in Lurie’s book with a pointer to the above entry.

• CommentRowNumber5.
• CommentTimeApr 25th 2013

I wish I wrote that reply, but I'm afraid that was Akhil (\ne Adeel!) ;)

• CommentRowNumber6.
• CommentAuthorUrs
• CommentTimeApr 25th 2013

Oh, sorry. I should be paying more attention, that’s embarrassing. Sorry for the confusion.

But anyway, thanks for the pointer!

• CommentRowNumber7.
• CommentTimeNov 20th 2014
• (edited Nov 20th 2014)

I added a remark, taken from an answer of David Carchedi on MO, about effective epimorphisms in sheaf toposes.

• CommentRowNumber8.
• CommentAuthorZhen Lin
• CommentTimeFeb 24th 2015

I recently had to be told that effective epimorphisms in the $(\infty, 1)$-category of spaces need not be epimorphisms. Perhaps a red herring principle warning is in order.

• CommentRowNumber9.
• CommentAuthorMike Shulman
• CommentTimeFeb 24th 2015

Yes, that’s actually already true in the (2,1)-category of groupoids. (Although I can’t remember whether I’ve ever heard someone use “epimorphism” to mean “monomorphism in the opposite category” for 2-categories or (∞,1)-categories.) Feel free to add.

• CommentRowNumber10.
• CommentAuthorZhen Lin
• CommentTimeFeb 25th 2015

I added some remarks to that effect.

I guess your example in 1-groupoids is $S^0 \to \Delta^0$?

• CommentRowNumber11.
• CommentAuthorDavidRoberts
• CommentTimeFeb 25th 2015
• (edited Feb 25th 2015)

Can we have some concrete statement other than ’it’s not true’? Or rather, what definition of ’epimorphism’ are you using?

• CommentRowNumber12.
• CommentAuthorUrs
• CommentTimeFeb 25th 2015

I have moved Zhen Lin’s addition to a numbered example and added a hyperlink to epimorphism in an (infinity,1)-category in order to clarify what is meant. Also added more cross-links there.

This concept of epimorphism in an $\infty$-category is rarely used, isn’t it.

• CommentRowNumber13.
• CommentAuthorCharles Rezk
• CommentTimeFeb 25th 2015

It does seem rarely used, though there are some nifty examples:

• A map $A\to B$ of commutative ring spectra is an epimorphism iff $B$ is smashing over $A$, i.e., if $B\wedge_A B\approx B$.

• A map $X\to Y$ between connected spaces is an epimorphism iff $Y$ is formed via a Quillen-plus construction from a perfect normal subgroup of $\pi_1 X$.

I sometimes try to find useful criteria for “epimorphism” in other settings. It’s usually pretty hard.

• CommentRowNumber14.
• CommentAuthorUrs
• CommentTimeFeb 25th 2015
• CommentRowNumber15.
• CommentTimeFeb 25th 2015
• (edited Feb 25th 2015)

A map $A\to B$ of commutative ring spectra is an epimorphism iff $B$ is smashing over $A$, i.e., if $B\wedge_A B\approx B$.

Isn’t it true that in any category admitting fibred coproducts, $f : x \to y$ is an epimorphism iff the codiagonal morphism $x \coprod_y x \to x$ is an isomorphism (and dually for monomorphisms)? Is the same true for $(\infty,1)$-categories? (The above would then just be a special case of this.)

• CommentRowNumber16.
• CommentAuthorZhen Lin
• CommentTimeFeb 25th 2015

That’s right. That is why suspension shows up in the (counter)example: for a space $X$, $X \to \Delta^0$ is an epimorphism in the $(\infty, 1)$-category of spaces if and only if the (unreduced) suspension $\Delta^0 \amalg_X \Delta^0$ is contractible. More generally, it seems to me that $X \to Y$ is an epimorphism in the $(\infty, 1)$-category of spaces if and only if its homotopy fibres are spaces with contractible suspension.

• CommentRowNumber17.
• CommentAuthorCharles Rezk
• CommentTimeFeb 26th 2015

Of course. The interesting feature of the commutative ring case is that the pushout is computed as a smash/tensor product, so whether $A\to B$ is an epimorphism of rings can be detected without appealing to the ring structures, and merely depends on $B$ as an $A$-module.

It goes the other way, of course: if $A$ is a commutative ring and $A\to B$ is a map of $A$-modules such that $B\approx B\wedge_A A\to B\wedge_A B$ is an equivalence, then $B$ is uniquely a commutative $A$-algebra, and $A\to B$ an epimorphism.

Another amusing fact: you can define a “Quillen plus-construction” of a commutative ring spectrum $A$, in complete analogy with the construction for spaces. Instead of killing a perfect subgroup of the fundamental group, the input data is a “perfect ideal” in the homotopy category of compact $A$-modules. All homotopy epimorphisms of commutative rings can be obtained this way.

People have discussed plus-constructions in other contexts (dg Lie algebras, for instance). These should probably give other examples of epimorphisms.

• CommentRowNumber18.
• CommentAuthorMike Shulman
• CommentTimeFeb 26th 2015

Is there a general $\infty$-categorical notion of “plus construction”?

• CommentRowNumber19.
• CommentAuthorCharles Rezk
• CommentTimeFeb 27th 2015
• (edited Feb 27th 2015)

I don’t know. By “plus construction”, I (approximately) mean a two step process where you (1) kill some stuff by introducing some “relations” $\{r_i\}_{i\in I}$, then (2) kill some more stuff by introducing some “higher relations” $\{s_i\}_{i\in I}$, where “relations” and “higher relations” are indexed by the same $I$. Furthermore, each “higher relation” should correspond to some kind of “redundancy” inherent in killing the “relations”.

For instance, given a space $X$ and a subgroup $P\subseteq \pi_1X$ generated by commutators $c_i=[x_i,y_i]$ of loops $x_i,y_i\in \Omega X$ representing elements of $P$, step (1) is: attach a 2-cell $d_i$ along each $c_i$, obtaining a space $Y$, while step (2) is: attach a 3-cell $e_i$ along the 2-sphere in $Y$ whose southern hemisphere is $d_i$, and whose northern hemisphere is $[H_i,K_i]$, built from choices of null-homotopies $H_i,K_i$ of the loops $x_i,y_i$ (which exist in $Y$ exactly because $P$ is generated by commutators).

The resulting map $X\to Z$ is an epimorphism.

Proof: The construction depends on the collection of choices $\alpha=\{(x_i,y_i,H_i,K_i)\}$ (assume fixed indexing set $I$), which themselves form a space $A$, and the plus-construction depends “continuously” on $\alpha\in A$. If $x_i$ and $y_i$ are themselves null-homotopic, then you can connect $\alpha$ to $\alpha_0=\{(*,*,*,*)\}$ (all constant maps) by a path in $A$, and it’s clear that the plus-construction built from $\alpha_0$ admits a deformation retraction, from which we conclude that $f$ is an equivalence when $[x_i],[y_i]$ are trivial in $\pi_1X$.

Next note that if $g\colon X\to X'$ is a map, then the pushout along $g$ of a plus construction $f\colon X\to Z$ built from an $\alpha$ is a map $g'\colon X'\to Z'$ which is itself a plus-construction built from $g(\alpha)$. It is clear that the plus construction map $f$ kills the elements $[x_i],[y_i]\in \pi_1X$, so the pushout of $f$ along itself must be an equivalence.

I don’t know too many other examples of this type of thing.

• CommentRowNumber20.
• CommentAuthorMike Shulman
• CommentTimeFeb 27th 2015

Very cool! I bet this has a nice formalization using HITs. However I don’t quite follow this bit:

If $x_i$ and $y_i$ are themselves null-homotopic, then you can connect $\alpha$ to $\alpha_0=\{(*,*,*,*)\}$ (all constant maps) by a path in $A$.

I see that you can connect $\alpha$ to something of the form $\{(\ast,\ast,H_i',K_i')\}$, but the constant loop can be nullhomotopic in a nontrivial way, so how do you know that $H_i'$ and $K_i'$ are also trivial?

• CommentRowNumber21.
• CommentAuthorCharles Rezk
• CommentTimeFeb 28th 2015

Whoops. I don’t. The real argument is: if $x_i=*=y_i$, the map on the “northern hemisphere” factors through a map $[H_i,K_i]\colon S^2\to Y$, which is null homotopic because $\pi_2$ is abelian.

1. added reference to HTT 6.2.3.10

Shane

• CommentRowNumber23.
• CommentAuthorMike Shulman
• CommentTimeJul 10th 2021

I wonder if comments 13-21 in this thread could be moved to a discussion thread for epimorphism in an (infinity,1)-category.

• CommentRowNumber24.
• CommentAuthorUrs
• CommentTimeJul 10th 2021