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    • CommentRowNumber1.
    • CommentAuthorFinnLawler
    • CommentTimeMay 3rd 2011

    The page constant morphism says

    As with Set, any morphism which factors through a terminal object is constant but although this is an “if and only if” in Set it need not be in a general category.

    I think the two are equivalent in any category with a terminal object, as explained here. That is, a morphism kk is constant if and only if the natural transformation k *k_* factors through the terminal object in the presheaf category, so that if the latter is representable then so is the factorization of k *k_*.

    Have I missed something? If not I’ll edit the page.

    • CommentRowNumber2.
    • CommentAuthorAndrew Stacey
    • CommentTimeMay 3rd 2011

    The original statement dates back to me and therefore is almost certainly me “playing safe”. Toby, Mike, and Urs all at least read the page and didn’t see fit to correct it, but they may have assumed that I knew what I was talking about (dangerous assumption here!).

    • CommentRowNumber3.
    • CommentAuthorUrs
    • CommentTimeMay 3rd 2011
    • (edited May 3rd 2011)

    I’ve only looked at it for a minute, but what FInn says looks right.

    (I have also done my standard editorial thing to the entry now… ;-)

    • CommentRowNumber4.
    • CommentAuthorUrs
    • CommentTimeMay 3rd 2011
    • (edited May 3rd 2011)

    didn’t see fit to correct it

    Wait, I don’t think there is anything outright wrong in the entry. The entry seems to say that not in every category does a constant morphism factor through a terminal object.

    What Finn observes is that if a terminal object exists, then every constant morphism factors through it. But a terminal object may not exist.

    • CommentRowNumber5.
    • CommentAuthorFinnLawler
    • CommentTimeMay 3rd 2011

    OK, thanks, that all mirrors what I suspected. I’ve edited the page now.

    • CommentRowNumber6.
    • CommentAuthorUrs
    • CommentTimeMay 3rd 2011

    Do we have an example of a constant morphism in a category without terminal object?

    • CommentRowNumber7.
    • CommentAuthorFinnLawler
    • CommentTimeMay 3rd 2011
    • (edited May 3rd 2011)

    Isn’t it true that the category of fields doesn’t have a terminal object? I saw this proved somewhere but I forget how. Then any constant morphism at 0 will be an example.

    Edit: no, that can’t be right, because if there was a field morphism with constant value 0 that would make 0=1 in the codomain.

    • CommentRowNumber8.
    • CommentAuthorMike Shulman
    • CommentTimeMay 3rd 2011

    Wait a minute, I don’t agree with the change. I think the identity morphism of any subterminal object is a constant morphism, but not every subterminal object admits a global section.

    • CommentRowNumber9.
    • CommentAuthorFinnLawler
    • CommentTimeMay 4th 2011

    That sounds plausible, but it seems to mean that the two aren’t equivalent even in Set, right?

    • CommentRowNumber10.
    • CommentAuthorMike Shulman
    • CommentTimeMay 4th 2011

    Yes, that’s right! The identity map of the empty set is constant according to this definition, but it doesn’t factor through the terminal object.

    That makes me less sure that this is the right definition of “constant morphism”.

    • CommentRowNumber11.
    • CommentAuthorFinnLawler
    • CommentTimeMay 4th 2011

    Well, that’s a bit of a bugger. I wonder if there’s a way to characterise the objects for which the equivalence fails; maybe it holds for morphisms out of non-subterminals or something. But it’s too late here to start thinking about it now.

    • CommentRowNumber12.
    • CommentAuthorMike Shulman
    • CommentTimeMay 4th 2011

    I think the equivalence holds for any morphism whose domain admits a global section. Proof: if f:XYf\colon X\to Y is constant as defined on the page, and XX admits a global section x:1Xx\colon 1\to X, then 1 X1_X and x!x \circ ! are two morphisms XXX\to X, hence f=f1 X=fx!f = f 1_X = f x \circ !, so ff factors through the terminal object 11. The converse is obvious.

    I think non-subterminals isn’t strong enough; if UU is subterminal, nonterminal, and nonempty, then the fold map U+UUU+U \to U is constant, but doesn’t factor through the terminal object. Similarly for the fold into one summand U+UU+UU+U \to U+U, and U+UU+U is not generally subterminal.

    • CommentRowNumber13.
    • CommentAuthorUrs
    • CommentTimeMay 4th 2011

    I think the equivalence holds for any morphism whose domain admits a global section. Proof:

    With that clause Finn’s previous argument is patched: the argument secretly assumed that the hom-spaces are inhabited in the first place.

    So it seems that with his argument the assumption can be relaxed to: assume Hom(K,X)Hom(K,X) is inhabited for all KK.

    • CommentRowNumber14.
    • CommentAuthorFinnLawler
    • CommentTimeMay 4th 2011

    Thanks, everyone. I’ve edited the page constant morphism accordingly.

    • CommentRowNumber15.
    • CommentAuthorMike Shulman
    • CommentTimeMay 4th 2011

    I don’t quite see how the two definitions can be equivalent in the absence of a terminal object, since in that case the “factors through the terminal object” definition doesn’t even make sense. And of course in the presence of a terminal object, Hom(K,X) being inhabited for all K is equivalent to X admitting a global section.

    • CommentRowNumber16.
    • CommentAuthorFinnLawler
    • CommentTimeMay 4th 2011

    I’ve clarified the phrasing a little.

    • CommentRowNumber17.
    • CommentAuthorTobyBartels
    • CommentTimeMay 4th 2011

    The empty map is constant in that it doesn’t vary, but it has no constant value. They’re different things, like the different senses of connected.

    • CommentRowNumber18.
    • CommentAuthorTobyBartels
    • CommentTimeMay 4th 2011

    I’ve put in a version of the strict definition that needs no terminal object.

    • CommentRowNumber19.
    • CommentAuthorMike Shulman
    • CommentTimeMay 4th 2011

    Nice, thanks Toby! That helps clarify. It really is just like the question of whether the empty set is connected.

    • CommentRowNumber20.
    • CommentAuthorFinnLawler
    • CommentTimeMay 5th 2011

    Somehow Toby’s edit restored the bit that Mike complained about in #15, so I’ve rewritten it again.

    • CommentRowNumber21.
    • CommentAuthorMike Shulman
    • CommentTimeMay 5th 2011

    I thought that at first too, but actually, Toby also changed the second definition so that that bit becomes true in a category without a terminal object. So I reverted your edit. (-:

    • CommentRowNumber22.
    • CommentAuthorFinnLawler
    • CommentTimeMay 5th 2011

    Oh, right, I didn’t see that bit.

    • CommentRowNumber23.
    • CommentAuthorSridharRamesh
    • CommentTimeMay 6th 2011
    In reply to Mike's "I don't quite see how the two definitions can be equivalent in the absence of a terminal object, since in that case the "factors through the terminal object" definition doesn't even make sense.":

    Presumably, the right way to interpret "factors through the terminal object" in categories without terminal objects is by passing to presheaves. Just as a morphism is constant (by definition #1) just in case its image under the covariant Yoneda embedding factors through a subterminal presheaf, so we might say a morphism is constant (by definition #2) just in case its image under the covariant Yoneda embedding factors through the terminal presheaf, regardless of whether the terminal presheaf is actually representable.
    • CommentRowNumber24.
    • CommentAuthorSridharRamesh
    • CommentTimeMay 6th 2011
    Whoops, I see now that this was already basically addressed.
    • CommentRowNumber25.
    • CommentAuthorMike Shulman
    • CommentTimeJul 25th 2013

    In case anyone is interested, essentially the same issue discussed earlier in this thread has come up recently on the homotopy type theory mailing list. In that context there are at least three proposed definitions of “constant function” f:XYf:X\to Y:

    1. for all x 1,x 2:Xx_1,x_2:X we have f(x 1)=f(x 2)f(x_1)=f(x_2).
    2. ff factors through the (1)(-1)-truncation X\Vert X\Vert of XX.
    3. there exists y 0:Yy_0:Y such that f(x)=y 0f(x)=y_0 for all x:Xx:X. Equivalently, ff factors through the terminal object, which is the (2)(-2)-truncation of XX.

    If YY is an hset, then (1) and (2) are equivalent, but even then both are weaker than (3) (e.g. when X=Y=X=Y=\emptyset); in this case we recover the dichotomy discussed at constant morphism. But when YY is not an hset, then (1) and (2) can also differ, e.g. there are models with f:XYf:X\to Y satisfying (1) where XX is a merely-inhabited set (i.e. its (1)(-1)-truncation X\Vert X\Vert is contractible) and YY is a 1-type with no global points, so there are no functions XY{\Vert X\Vert}\to Y.

    For that reason, I’m inclined not to consider (1) as a notion of “constant function”, but I’m curious to hear other opinions.

    • CommentRowNumber26.
    • CommentAuthorZhen Lin
    • CommentTimeJul 25th 2013

    It looks to me that (1) is deficient because it comes from the truncation of a hypercover, no? At least, here is how I think about it in the 1-topos case: (1) says that ff factors through the coequaliser of the two projections X×XXX \times X \to X, which is precisely the image of X1X \to 1.

    • CommentRowNumber27.
    • CommentAuthorMike Shulman
    • CommentTimeJul 25th 2013

    Yes, I entirely agree with you. Unfortunately, I haven’t managed to convince Martin Escardo of that yet. Care to help?

    • CommentRowNumber28.
    • CommentAuthorZhen Lin
    • CommentTimeJul 26th 2013

    I tried reading through the preceding discussion but I’m afraid I can’t really follow…

    • CommentRowNumber29.
    • CommentAuthorMike Shulman
    • CommentTimeJul 26th 2013

    Sorry. It is a bit all over the place.

    • CommentRowNumber30.
    • CommentAuthorTobyBartels
    • CommentTimeAug 10th 2013

    I don't understand this bit:

    YY is a 1-type with no global points

    I read this as

    YY is a groupoid with no objects

    which doesn't work.

    But possibly I am mixing internal and external language. Is the resolution that you working in a non-well-pointed model of HoTT?

    In any case, I can follow the main argument, as I can distinguish (1) and (2) in a non-well-pointed topos, or in the category of locales, or in other places where non-empty things may have no global points.

    • CommentRowNumber31.
    • CommentAuthorMike Shulman
    • CommentTimeAug 10th 2013

    Yes, the phrase “global point” is only really meaningful at the meta-level. In type theory the appropriate statement is that there is no closed term of type Y, which is a meta-statement about the type theory, and distinct from the internal assertion “Y is not inhabited”.

    • CommentRowNumber32.
    • CommentAuthorMike Shulman
    • CommentTimeJul 6th 2016

    Prompted by this MO question I looked back at the page constant morphism, and I don’t think the terminal-object-free version of Definition 2 is correct. If, as suggested by Sridhar and Qiaochu, we define a morphism to be constant if its image under the Yoneda embedding factors through the terminal presheaf, then I think we would get something like “we can choose for each object XX a morphism f X:XCf_X : X \to C such that for any g:YXg:Y\to X we have f Xg=f Yf_X g = f_Y and moreover for any h:XBh:X\to B we have ch=f Xc h = f_X”. In other words, for each object AA there is a specified morphism f A:ACf_A:A\to C, chosen naturally in AA, such that if any morphism ACA\to C factors through cc then f Af_A is the unique one that does. This includes, for instance, the map 010\to 1 in SetSet (which factors through the terminal object), whereas the definition as stated does not.

    • CommentRowNumber33.
    • CommentAuthorSridharRamesh
    • CommentTimeJul 7th 2016
    • (edited Jul 7th 2016)

    It seems to me we can say more simply that c:BCc : B \to C is constant, in the sense corresponding to factoring through 11 even if no terminal object is actually around, just in case “we can choose for each object XX a morphism f X:XCf_X : X \to C such that for any g:YXg:Y\to X we have f Xg=f Yf_X g = f_Y and moreover we have c=f Bc = f_B”.

    • CommentRowNumber34.
    • CommentAuthorMike Shulman
    • CommentTimeJul 8th 2016

    I think you’re right. I’ll fix the page when I get a moment, unless someone else has done it first.

    • CommentRowNumber35.
    • CommentAuthorMike Shulman
    • CommentTimeJul 8th 2016

    Fixed.

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