I've always wanted to learn about topology. Your video playlist and the Covid confinement gave me a great opportunity :) Thanks again!
@rtg_onefourtwoeightfiveseven4 жыл бұрын
31:40 1:05:00 I'm not convinced that the examples he's giving are manifolds. For the first one it looks like it's not locally homeomorphic to R^2 at point q, and for the second the dimension of any open set that contains the boundary seemingly isn't consistent - it's homeomorphic to R^(dim M + 2) to the left and R^(dim M + 1) to the right.
@yongminpark59097 жыл бұрын
Wow, what a remarkable and fascinating explanation about the chart representation of a continuous curve! I haven't seen before like this. With aid of this lecture, I can completely understand the purpose of manifold concept and its usefulness.
@uttarandutta43764 жыл бұрын
I am interested in algebraic geometry and want to go into schemes. So I wanted to learn manifolds and vector bundles etc properly. Well I stumbled upon this series, and now all of a sudden I want to devote my full energy to learning manifolds. Thanks youtube for this recommendation. Professor, you have become one of my favorites. Great job.
@ark89499 жыл бұрын
these are wonderful lectures, they have the best discussion of fibre bundles I have seen
@tim-701cca8 ай бұрын
summarize some mistakes: 9:20 It should be S1 and figure 8 is not homeo. to circle 32:43 E is not a manifold 50:58 (u,f) continuous 57:20 Mobius strip and cylinder are not homeo. and isomorphic as bundle 1:00:36 restricted bundle is bundle isomorphic to some "subbundle" of (E′, π′, M′) 1:05:00 E is not a manifold
@KushagraSachanIITBHU5 ай бұрын
Btw, at 9:20, that's not a figure 8. More like the perimeter of a Pringles chip. Thanks for the summary though!
@丁伟-l4s6 жыл бұрын
I love these so much, I have looking for these stuff explaining connection between math and physics which make me so clear for so long. Thanks !
@maxwang51097 жыл бұрын
This is my favorite kind of Biology.
@UnforsakenXII7 жыл бұрын
By the way, when are we going to the gym tomorrow? LOL.
@maxwang51097 жыл бұрын
Lmao whenever
@redaabakhti7683 жыл бұрын
Wait you said tomorrow but its the end of the universe let postpone to yesterday (makes sense if time is cyclic)
@redaabakhti7683 жыл бұрын
Makes sense if noting makes sense
@ericgamliel85007 жыл бұрын
These lectures are great. His mistakes are (mostly) obvious and forgivable.
@eastwestcoastkid4 жыл бұрын
What mistake has he made?
@brani1144 жыл бұрын
@@eastwestcoastkid At 57:15 he said for example that the cylinder was homeomorphic to the möbius strip, which is not true.
@eastwestcoastkid4 жыл бұрын
@@brani114 he said as fibre bundles they were not homeomorphic which is correct they’re not-but he should have stated that’s true for the total space as well-is that what you’re saying?
@brani1144 жыл бұрын
@@eastwestcoastkid Yes, he was trying to give an example of non isomorphic fiber bundles that have homeomorphic total spaces and homeomorphic base spaces. But the möbius strip/cylinder example is just not a good example, since the total spaces are not homeomorphic.
@brani1144 жыл бұрын
@@eastwestcoastkid A better example would have been: The circle once as fiber bundle over itself by the identity mapping and once as fiber bundle over itself by the mapping z^2 (the circle covering itself twice).
@YourFriendlyAlan Жыл бұрын
1:00:36 There is a mistake in the definition of locally isomorphic bundles. Rather, the restricted bundle should be isomorphic to the subbundle (u(preim_π(U)), π′|_u(preim_π(U)), f(U)) of (E′, π′, M′) where (u,f) is a bundle isomorphism. The fact that (u,f) is a bundle isomorphism means that u and f are assumed to be homeomorphisms with the relevant commutative property. In this way, f(U) is a neighbourhood of f(p) just as U is a neighbourhood of p and moreover, these neighbourhoods are isomorphic. It is also clear that (u(preim_π(U)), π′|_u(preim_π(U)), f(U)) is indeed a subbundle of (E′, π′, M′) by definition of u and f. Hence, this definition makes sense. Intuitively, one can see that this definition makes sense since if (E, π, M) is the cylinder and (E′, π′, M′) is the Möbius strip then they are clearly locally isomorphic bundles. On the other hand, unfortunately, Schuller’s given definition does not make sense as this would imply that the cylinder and the Möbius strip are not locally isomorphic as there is no subbundle of the cylinder that is isomorphic the Möbius strip. Actually, now that I’ve written this, maybe a cleaner definition should be (E, π, M) and (E′, π′, M′) are locally isomorphic bundles if for every point in E there is an open neighbourhood of p such that the restricted bundle is bundle isomorphic to some subbundle of (E′, π′, M′). Ah, yes, the words “some subbundle of (E′, π′, M′)” would have saved the day here. Enjoy my thought processes KZbinrs!
@xrhsthsuserxrhsths6 жыл бұрын
In the example at 33:40 i think that E is not a manifold because at the point q its dimension is 1 while everywhere else it's 2
@curtjaimungal6 жыл бұрын
This is what I was thinking. Except that q is dim 0 and the rest are dim 1. Aren't manifolds supposed to have constant dimension, which is why the notion of dimension is well-defined on a manifold by definition?
@jackozeehakkjuz5 жыл бұрын
Came here to say this. You are exactly right.
@jackozeehakkjuz5 жыл бұрын
Also the example given at 1:04 is not a bundle of topological manifolds, because at some points it is of real dimension 3 and in others it is of real dimension 2. This is clarified in lecture 10, at 1:32:22.
@billf75853 жыл бұрын
@@curtjaimungal The rest are indeed dim 2, locally around p the manifold looks like R x S1 , so the surface of a cylinder which is dim 2. And for q its more accurate to say there is no meaningful dimension, since there is no open neighborhood which maps homeomorphically to R^d for any d. Either way, you are both right, the example fails to be a manifold. Edit: I drew a picture for fun imgur.com/a/fRGBHA9 its like a strip of paper pinned to the tip of a cone.
@andreemcaldas6 жыл бұрын
I believe the cylinder and the Möebius strip are not homeomorphic. (57:20)
@kockarthur79766 жыл бұрын
Correct, I think that was just a slip of the tongue. He probably meant to say that although, when regarded as fiber bundles, the Mobius strip and cylinder have homeomorphic base spaces and typical fibers, they are still not isomorphic as fiber bundles.
@p0gr4 жыл бұрын
came here to say that.
@cartmansuperstar2 жыл бұрын
how can one reason that ?
@MsSlash892 жыл бұрын
@@cartmansuperstar Usually, to prove that two topological spaces are not homeomorphic, you look for a property that remains invariant under isomorphism (i.e. all homeomorphic spaces have it equal) and notice that for a specific pair of spaces it is actually different. The boundary of the cylinder consists of two disjoint circles, top and bottom; the boundary of a Möbius Strip is actually homeomorphic to a single circle. If they were homeomorphic, their boundaries would be homeomorphic.
@dr.leonardhofstadter58668 жыл бұрын
Great lectures on topology, i like the fact that he exaplains, the topological spaces. One of my favorite subjects,
@mappingtheshit3 жыл бұрын
It is indeed a great fact that he is teaching topology. The fact... U r not bright, aren’t u? U think watching a sitcom can make u a physicist? Dumb idiot
@jovaha8 жыл бұрын
Isn't the S^2 at 8:26 supposed to be S^1?
@jackozeehakkjuz7 жыл бұрын
Yes, it should be.
@mappingtheshit6 жыл бұрын
no, you are wrong... the interior is not empty
@rewtnode6 жыл бұрын
General of your mom that may be true for the circle and the square he drew, but for last object, which looked like a loop in space, he didn’t make any indications about it being somehow filled. The loop itself is definitely S1. If that thing is supposed to be S2 it ought to be drawn different so there is an interior visible.
@tomkite19335 жыл бұрын
@Bennabi Mehdi If it is supposed to be a disk then it is a different set usually called D^2. It has the same definition as the S sets but change the x^2 + y^2 =1 to x^2 + y^2
@hershyfishman29295 ай бұрын
@@mappingtheshit he clearly says at 10:24 that locally it is part of the real line, so he means s1
@yigal_s7 жыл бұрын
I do not see how a total space may be a manifold if its fibers are not homeomorphic. Particularly, is the space on 32:13 a manifold? It is not locally homeomorphyc to R^2 around point q.
@neildhan8 жыл бұрын
I forwarded straight to the part on bundles, and was very impressed with your lecture. As Amara Katabarwa said below, this is the best discussion I have seen on the subject. I didn't have time to watch all of it, but I have liked and subscribed, and will definitely be back. I'm sure you'll have much more that will be of help to me. Thanks!
@vivalibertasergovivitelibe41112 жыл бұрын
This is an awesome lecture to revisit topological bundles. Thanks!
@zfengjoe3 жыл бұрын
At 1:44:24, I think the reason of the original r is continuous if xr is continuous is that x is a homomorphism. In general, x being only a continuous map is not enough (unless the topology on M is as weak as the weak-topology induced by x).
Question: The definition of a topological manifold as presented here does not include manifolds with boundary. So, the Mobius strip should not be a manifold as points on the boundary do not have an open neighbourhood homeomorphic to R^n, right? Or we can take the boundary to be open, which would give the interval (-1, 1) at 29:38.
@Nico-vj7jc4 жыл бұрын
29:59 This is not a manifold, but a manifold with boundary. For points on the upper edge of the square won't have euclidean neighbourhoods (after identification)
@brendawilliams80622 жыл бұрын
I am glad he clarified the Möbius strip is not a product manifold.
@KipIngram3 жыл бұрын
57:50 - Why is the projection function different for the cylinder and the Moebius strip? I thought it just pointed at the base space.
@brandonwillnecker80603 жыл бұрын
The cylinder is understood as vertical intervals attached around the circle. The projection would then be : take the "above" points and map them down to the circle, and take the "below points and map them to the circle. This is the same for all points. However, for the moebius strip we have that the "above" points end up becoming the "below" points during the twist. Think of the paper model. So now this change in placement needs to be taken care of in the projection mapping in a continuous way..again thinking of the continuous twisting of the paper model.
@snaqvi697 жыл бұрын
Great lecture! Is it supposed to be S1(not S2) at 8:30?
@mappingtheshit6 жыл бұрын
no, you are wrong... the interior is not empty
@bendonahoo85635 жыл бұрын
Shahid Naqvi Yes
@BobSaget-et6ln3 жыл бұрын
Yes. S^1 is the "circle".
@billf75853 жыл бұрын
@@mappingtheshit Well that doesn't make sense, if the interior were not empty it would be D2 not S2, surely he meant S1.
@netrapture2 жыл бұрын
He probably meant S^1 is homeo to the square('s bounday) is homeo to the (boundary curve) of the wavy disc embedded in R^3, esp since later he says it is not homeo to R because it is compact, like he's assuming the same dimension, otherwise S^2 would be even more obviously nonhomeo to R on basis of dimension.
@michellejingdong Жыл бұрын
Thank you so much! This is so clear and helpful!
@ahmeddalile6 жыл бұрын
1:44:44 "Imagine a Bird flying through this Room" : * "Room" is isomorphic to "Theoretical Physics" embedded in "KZbin". * "Bird" is isomorphic to "Dr F. Schuller".
@ryanjbuchanan2 жыл бұрын
Where can I find the problem sheets for this lecture?
@volcanic31042 жыл бұрын
Can someone explain how E at 32:18 would constitute a manifold? To the left of q, any point on those circles would be locally homeomorphic to R except for the point that joins the circle to the real line, which would the homeomorphic to R^2. On q, it looks like R and then to the right of q, again we have points that locally look like R^2. The dimension does not appear to be well-defined...
@SirTravelMuffin Жыл бұрын
I completely agree! Was wondering that myself as I was reviewing the notes. Since the total space needs to be a manifold, I think this example would technically not be a bundle. I wonder if the fibers need to be homeomorphic even if it is not a fiber bundle in order for the total space to be a manifold. Have you had any more thoughts on this?
@Spykoni1 Жыл бұрын
Has anybody understood (42:15 - 45:55) why the wave function is not a function, but rather a section over a C-bundle? Isn't it the trivial product CxR^3? Why you need the concept of bundle here? Probably the answer is that a wave function is defined modulo a phase (a phase is not an observable), hence there is this phase arbitrariness of the "wave functions" that make them not actually behave like real functions, but I'm not really sure about that.
@netrapture Жыл бұрын
See this later lecture kzbin.info/www/bejne/eWqWfK2AbJJ4qZY
@JaechulLee-u2k Жыл бұрын
31:13 The bundles are the generalization of the idea of taking a product.
@alexbaykov92217 жыл бұрын
I'm confused. Shouldn't it be like product bundles (not mfds) < fibre bundles < bundles?
@kockarthur79766 жыл бұрын
Yes, but he has in the back of his mind physics where pretty much all topological spaces of interest are manifolds.
@iamyouu5 жыл бұрын
Can someone please post the problem sheet, please...!
I guess the space E in example at 31:40 is not a topological manifold for the same reason the "cone" isn't.
@andreemcaldas7 жыл бұрын
Also, at 1:05:00.
@aeroscience98346 жыл бұрын
Andre Caldas it's not a manifold if you take his drawing as being a subset of R^2 with the induced topology, but that's not what he was trying to illustrate.
@rudrayangooptu94762 жыл бұрын
Had a small doubt(might be trivial) Is the union of the fibres(as sets) for all the points in the base space equal to the total space?
@kvazau84442 жыл бұрын
yes, by the fact that the projection map is defined on all of the total space
@hemantamandal65125 жыл бұрын
Wish my teachers were this great
@rewtnode6 жыл бұрын
40:20 when he defines the section of a bundle his explanation is clear but leaves me still puzzled why this construction is called a section? I suppose it makes sense when I consider some kind of line connecting all the points in E that are the images of Sigma applied to the base space M in his sketch.
@alexsiryj6 жыл бұрын
Sections are also sometimes called cross-sections. They generalize the notion of the graph of a function. So for example in R^2, with the base space being the x-axis and the fibers being the y-axis, we define the standard fiber bundle with projection pi(x, y)->x. With this definition, a section is any function of the form sigma(x)=(x,f(x)), which is exactly how we're used to plotting the graph of a function. Proving that sigma is a section is easy since pi(sigma(x))=pi(x,f(x))=x which is the identity function. It gets more interesting when the fiber bundle is twisted or distorted in some way, allowing us to "graph" sections along the base space similar to how we graph functions in R^2.
@yogeeshreddy63735 жыл бұрын
@@alexsiryj Ah! Now I can appreciate it better. Thank you
@Nico-vj7jc4 жыл бұрын
That's a naming convention from category theory. Just search for "section category theory".
@Nico-vj7jc4 жыл бұрын
Also there is a good explanation for this name in the book "conceptual mathematics" in the context of Set.
@abrlim55972 жыл бұрын
In 30:13, must the Mobius strip be the result of this construction? Why can't a cylinder be a result of this construction?
@jimallysonnevado39733 жыл бұрын
Shouldn't the total space be also a manifold? Why the example in ~31:50 the total space E does not seem to be a manifold because it somehow has intersections?
@cantcommute2 жыл бұрын
it isn't intersecting he's drawing an illustration of the fibres of the points (turns out not to be a manifold, though, because at q your local open set is not homomorphic to R^d but to R^d-1)
@lemniscatepower3153 Жыл бұрын
at 1:16:50 … Is it like this, pullback of section = u^-1(sigma(f(m'))) . Can anyone please suggest some correction if it's wrong. Thank you.
@brennerlattin Жыл бұрын
I think you're close. We want the pullback of the section to be a section itself, meaning it takes on values in E', but u^-1 can give us a set. For example, if m1' and m2' are both mapped by f to the same value in M (that is, f(m1') = f(m2') = m for some m in M) then u^-1(sigma(f(m1'))) = u^-1(sigma(f(m2'))) =u^-1(sigma(m)) must be a set containing (m1', sigma(m)) and (m2', sigma(m)) since both are elements of E' and u((m1', sigma(m))) = u((m2', sigma(m))) = sigma(m). What I think we want to do instead is let sigma' be defined by sigma'(m') = (m', sigma(f(m'))), which ensures that sigma' is a section even if f is not injective and that the graph commutes.
@seba4580210 ай бұрын
In 1:21:15 technically the components functions x^i it should be the composition map proj_i o x? because its input is an element of M.
@tim-701cca8 ай бұрын
Yes, he defines x^i as the composition.
@andreemcaldas6 жыл бұрын
In 49:21, how do you know that the pre image is in fact a manifold?
@kockarthur79766 жыл бұрын
He did not explain it, but in fact the fibers on a fibered manifold (which seems to be his construction) are indeed submanifolds of the total space.
@tim-701cca8 ай бұрын
His definition of fibre bundle is not enough to prove that. It should be restricted to locally trivial fiber bundle.
@mehmetsolgun68168 жыл бұрын
He mentioned in the last lecture that the cartesian product top. space of two paracompact top.spaces may not be paracompact. But in this lecture, he defined the product manifold as a cart. product of two top. mfld.s. In this sense product space may not be paracompact and so not a mfld.. :/ on the other hand, all manifolds are paracompact by his definition... I feel there is something wrong there:/
@alvaroballon71338 жыл бұрын
=( you're right! I guess that's why most authors consider manifolds to be second countable, or else the product manifold is well defined only if one of the factors is compact =/.
@YouWillDieByMyHand7 жыл бұрын
It is something like this: Locally Euclidean -> Locally Metrizable -> Metrizable (Smirnov metrization theorem) -> Product of two metrizable spaces is a metrizable space -> Every metrizable space is paracompact -> Product is paracompact.
@rounak51064 жыл бұрын
@8:40 He said that since a circle can be continuously deformed into a square, it is topologically homeomorphic. But, shouldn't that be homotopic as the latter is a weaker condition than former?
@00TheVman4 жыл бұрын
You are right he should have said that there is a bijective map continuous in both directions, a homeomorphism, between them. I think lots of people say "continuously deformed" when they mean homeomorphism. You should be careful however since for a homotopy the ambient space and how the two subsets are embedded is important. For example the square {(x, y) in R^2 | max(|x|, |y|) = 1} and the circle {(x, y) in R^2 | x^2 + y^2 = 1} are not homotopic inside R^2 minus the point (0.8, 0.8). In this sense saying "homotopic is a weaker condition than homeomorphic" as homotopy relies on an ambient space and how the sets are embedded while homeomorphic is just between the abstract topological spaces.
@tonymok77524 жыл бұрын
@@00TheVman when talking about homotopy, the circle and the square should be interpreted as curves in R2, isn't it?
@00TheVman4 жыл бұрын
@@tonymok7752 Yeah for sure, I guess I didn't make that clear and just gave the images of the curves. Maybe there is a sense in which sets rather than maps can be deformed, but I'm not aware of it. If I recall correctly there is possibly a metric on the collection of compact subsets of a metric space, but I don't know for sure.
@cocoaaa2680 Жыл бұрын
39:36 how is section diff from preimage of pi? is it a generalization? example ofna map from base top.nmanifold to a total top manifold that is a section but not a preimage?
@cocoaaa2680 Жыл бұрын
41:26 conditioning or marginalizing a joint prob. distribution is this special case
@cocoaaa2680 Жыл бұрын
42:00 section vs a function 53:00 isomorphic as bundles. defn. qs: what eiuld be a degn of equivalence. isimorphism btw two fynctions. in particular teo prob dists ove the ame fomain. 55:00 same E, same base manifd. but if the twisting is encoded doff via thevorojection then the two bundles are not isomorphic as bundles 58:30 defn of local isomorphic as a bundle qs: grn. models (kapping density function) is a surjective map?? 1:08:00 wave fuctíon. local tivialjty of a bundle 1:09:00 pull back bundle of a bundle. import. notion fefn and how to construct it can construct a morphism based on the constructed pullback bundle 1:18:00 swich from bundles to "viewing manifods from atlases"
@cocoaaa2680 Жыл бұрын
1:29:00
@cocoaaa2680 Жыл бұрын
c0 compTibility of two charts (at the sMe point p on top manifold)
@cocoaaa2680 Жыл бұрын
1:48:30 for new gan proposal
@OhadAsor9 жыл бұрын
best exposition ever
@comkunal3 жыл бұрын
Any good reference which closely follows this ?
@abrlim55973 жыл бұрын
What is complex line? How can complex numbers be arranged in a line?
@educationtarunramkanuri88763 жыл бұрын
I too have the same question.
@gfcortes15463 жыл бұрын
It's not a line in the sense of a continuum that can be given an ordering that is compatible with the field operations. This is obvious. But in this context, line is referring to the dimension of the fibers in a line bundle, where by definition a line is a topological space with dimension 1. And the vector space of complex numbers (with appropriate topology) is 1-dimensional over the field of complex numbers. Hence, a complex line, so to speak.
@ANSIcode9 жыл бұрын
The requirement is missing that the image of the coordinate map must be open in R^d, in the definition of manifold.
@krzysztofbielas43298 жыл бұрын
+ANSIcode x is homeomorphism :)
@andreemcaldas7 жыл бұрын
Homeomorphism over its image! In principle, the image needs not to be an open set.
@jamma2467 жыл бұрын
In the definition he gave _any_ subspace of R^d would count as a topological manifold (take each U as that subspace, and x as the identity).
@juliangarcia34164 жыл бұрын
What is a composition of a function and a curve gives? Example??
@brendawilliams80622 жыл бұрын
The end points of 375 will not agree with three, so it gives a twist that affects the mobius strip.
@abrlim55974 жыл бұрын
Hi professor, why do you have one def for topological manifold in this video different from the one you introduced in another video?
@noobiechess86643 жыл бұрын
The maps @50:55 need to be continuous
@jiaweiwu35052 жыл бұрын
Thanks. I was just wondering why he talks about homeomorphism in bundle isomorphism. If u,v are continuous, then it makes sense.
@jacquessmeets44272 ай бұрын
No, there is no need that U and V must be continuous. A bundle morphism is simply a different word for a bundle map. One could of course consider special cases where u and v are assumed to be continuous, or diffeomorphisms, or are smooth etc.
@rewtnode6 жыл бұрын
21:30 great lecture, but jeez, he can’t draw a Moebius strip. I begin to get an intuition why many mathematicians love extremely elaborate formal constructs. The funny thing is that at this point he gives a very good motivation for bundles. Just try to draw a Moebius strip as a product manifold - it doesn’t work.
@rewtnode8 жыл бұрын
This is a great lecture series. Not too hard to follow and very systematic. Still, it's so easy to forget the details - there are simply too many definitions in this field, and if I return a week later to this topic, in my mind much of it has turned to mush. What I'm missing are compelling drawings to illustrate many of the proofs. The symbols and fancy notation is necessary but somehow isn't good enough. Too rarely does he try to draw, but when he does, it's quite helpful. Is there a good straight forward book or lecture notes that would roughly follow the same material in this order?
@iamyouu5 жыл бұрын
Lets be honest forget about illustrations for these sort of abstract concepts. However Frankels book is good
@LeoHsieh4 жыл бұрын
Sir. Roger Penrose's book : "The road to Reality" I found his drawing is extremely good and expressive.
@TheNonHiddenSingularity15 күн бұрын
why the section is called sigma and not just pi^-1 ? maybe a naive question
@kuroshkabir1362 жыл бұрын
thanks
@kapoioBCS5 жыл бұрын
I dont get how we know that the for every p the preimage of {p} under π forms a manifold ? (in the case of the fiber bundle ) :S
@LadliyadavLado-pj8lh Жыл бұрын
Nice sir tq
@SphereofTime Жыл бұрын
8:23
@NawSai-q2o2 ай бұрын
1:26:46 U are certainly compatible with someone if u never meet that person lol
@KipIngram3 жыл бұрын
30:20 - No, I don't see. I understood all the things you just said, but it looks entirely arbitrary and pointless to me. Why is it of value? How can I put it into an application context that makes it clear what benefits its bringing to me? This is really the whole reason I watched this course up to now - to get an understanding of this whole "fiber bundle" business that I'm constantly running into in my studies. But I'm not getting any enlightenment here.
@kapoioBCS5 жыл бұрын
Also I think he didn't defined correctly what locally homeomorphic means for the manifold (in the definition for the manifold). Because locally homeomorphic is not equivalent to the existence of a local homeomorphism. Two spaces are homeomorphic iff there exists a homeomorphism between them, but it's NOT true that two spaces are locally homeomorphic iff there exists a local homeomorphism between them.
@ashwalls104 жыл бұрын
Frederic Schuller should change his name to Lehrer!
@mohammedtayebbenmoussa45243 жыл бұрын
Chat happening how
@serhiypidkuyko42063 жыл бұрын
The definition of submanifold given by the lecturer is not good
@YourFriendlyAlan Жыл бұрын
1:17:23 It took me far longer than I’m going to admit to realise σ′(m′):=(m′, σ(f(m′))).