I've only just “discovered” homotopic functions, so apologies if I'm asking nonsense. Shifrin in Multivariable Mathematics (p405) gives an example of a homotopic function as$$H\left(\mathbf{x},t\right)=\left(1-t\right)\mathbf{x},$$

where $f:D^{n}\rightarrow D^{n}$, $f\left(\mathbf{x}\right)=\mathbf{x}$ (here $D^{n}$ is the unit closed ball) and $g\left(\mathbf{x}\right)=0$. For my own curiousity, I'm trying to see how, for the simple example of $D^{2}$, $$\intop_{D^{2}}f^{*}\omega=\intop_{D^{2}}g^{*}\omega,$$which I imagine is blindingly obvious, but I just can't see how to proceed. I'm not even sure what $\omega$ to use (I believe $\omega$ needs to be a closed 2-form on $D^{2}$). Using $\omega=dx\wedge dy$, for the lhs I get$$\int_{-1}^{1}\int_{-\sqrt{1-y^{2}}}^{\sqrt{1-y^{2}}}dx\,dy=\pi,$$which can't be right as $$\intop_{D^{2}}g^{*}\omega$$ must be zero. Any hints?

EDIT 1. Sorry, I got my original “unit disk” integral wrong. Shifrin says that the pullback integral applies when $X$ is a compact, oriented k-dimensional manifold and $f,g:X\rightarrow$ Y are homotopic maps, and $\omega$ is any closed k-form on $Y$.

EDIT 2. Following from the comment by @Kajelad that this only works if $X$ is boundaryless, I thought I'd try another simple example, ie $f:S^{1}\rightarrow S^{1}$, $f\left(\theta\right)=\theta$ and $g\left(\theta\right)=0$ and try to show $$\intop_{S^{1}}f^{*}\omega=\intop_{S^{1}}g^{*}\omega.$$But, again, I'm not sure how to do this.

EDIT 3. My question has now evolved to simply whether Shifrin's Propostion 7.4 only applies where $X$ is boundryless. Here's Proposition 7.4:

$X$ is a compact, oriented k-dimensional manifold and $f,g:X\rightarrow Y$ are homotopic maps. Then for any closed k-form $\omega$ on $Y$, we have$$\intop_{X}f^{*}\omega=\intop_{X}g^{*}\omega.$$

I can't see where he has defined $X$ as boundryless, but I may have missed that.

An online reference that confirms that $X$ and $Y$ are boundaryless would be most helpful.

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    $\begingroup$ Why do you expect that $\int f^*\omega=\int g^*\omega$ here? This certainly does not hold in general. Also, it seems your double integral should evaluate to $\pi$. $\endgroup$
    – Kajelad
    Aug 22, 2020 at 9:15
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    $\begingroup$ Your lhs integral doesn't make much sense $\endgroup$
    – giobrach
    Aug 22, 2020 at 9:18
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    $\begingroup$ With regards to edit 2, the map $f(\theta) = \theta$ and $g(\theta) = 0$ are not homotopic as maps to $S^1$. (And what, exactly, does $g(\theta) = 0$ for a map $g:S^1\rightarrow S^1$ mean? What do you mean by $0\in S^1$?) $\endgroup$ Aug 24, 2020 at 13:04
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    $\begingroup$ Unfortunately, I don't own Shifrin's book. But a) he frequents this website, so he may see this question and b) If my experience "manifold" is defined as "locally homeomorphic to $\mathbb{R}^n$" (plus some more technical conditions, e.g., Hausdorff, Paracompact). Can you find where he defines manifolds in the book? My guess is that he does not allow manifolds to have boundary. (There is an english/grammar issue here. A "manifold with boundary" is not a manifold with extra structure (a boundary). Rather, a manifold is a special manifold with boundary: one whose boundary is empty.) $\endgroup$ Aug 24, 2020 at 17:46
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    $\begingroup$ And your example with the disk shows that the theorem is false without extra hypothesis (e.g. that "manifold" means "manifold without boundary", as is common.) $\endgroup$ Aug 24, 2020 at 17:51

2 Answers 2



  1. The theorem that pullbacks via homotopic maps give the same integral: that's for manifolds, not manifolds-with-boundary. (Where "manifolds" really means "manifolds without boundary". See below.)

  2. Your second example, with $S^1$, uses the function $f(\theta) = \theta$, g(\theta) = 0$, and presumably the homotopy

$$ H(\theta, s) = (1-s) \theta $$

Unfortunately, this doesn't actually work. I'm assuming that Ted defines $S^1$ as the manifold consisting of all the points $\{ (\cos t, \sin t) \mid 0 \le t < 2\pi \}$, with the topology arising from inclusion in $\Bbb R^2$. So what is $f$? $f$ is actually a map $[0, 2\pi] \to [0, 2\pi]$, not a map from $S^1$ to $S^1$. What's hidden here is a map $p$ that identifies a point $$ \theta \in [0, 2\pi) $$ with the point $$ (\cos \theta, \sin \theta) \in S^1. $$ Because $p$ is a bijection, we can "refer to points in $S^1$ by their polar coordinates," which is how $f$ got defined. But while the function $p$ is continuous, the inverse function $p^{-1}: S^1 \to [0, 2\pi)$ is discontinuous, which can be a problem. For the real function you want, from $S^1$ to $S^1$, is defined by taking a point of $S^1$, finding its polar coordinate (i.e., applying $p^{-1}$, applying $f$, and then applying $p$ to get back to the unit circle. In short, you're looking at $$ F = p \circ f \circ p^{-1}, $$ while talking about $f$. Similarly for $g$.

Now $H$, as written above, is a homotopy from $f$ to $g$, so you might hope that $$ K(u, s) = p( H(p^{-1}(u), s ) $$ would be a homotopy on $S^1$. (Here $u$ denotes an arbitrary point of $S^1$; we extract its polar coordinate, apply $H$, and then push back to the unit circle with $p$.)

If $K$ we defined by composing $H$ with some continuous functions, it really would be a homotopy between functions $S^1 \to S^1$. But $p^{-1}$ is discontinuous, so the function $K$ that you're implicitly using...you can't use a continuity argument to show it's a homotopy, and in fact, it's not actually a homotopy. So the thing you'd like to conclude, that $$ \int_0^2\pi F^{*}\omega = \int_0^{2\pi} G^{*} \omega \tag{1} $$ cannot be "proved" using the cited theorem, because the hypotheses of the theorem are not all satisfied. And that's a good thing, because, as you observed with your own computations, the two sides of the "equality" in Equation 1 are not, in fact, equal.

In short: there's no contradiction to the claim's in Shifrin's book.

Extra note: The whole manifold-without-boundary thing can confuse people, and rightly so. If I say I have a blue truck, people tend to assume I have a truck, and that it's a little special: it's blue. If I have a friend-with-benefits, you assume that I have a friend. But if I say I have a manifold-with-boundary, you're not allowed to assume I have a manifold? What's up with that?

Well...historically, people started talking about manifolds first -- things like a sphere or a torus, etc. And then they realized that they wanted to slightly generalize this, and sometimes allow things like a disk, or a cylinder $S^1 \times [0, 1]$ into the same discussions. These things were not literally manifolds, but they had a LOT of the same properties, so someone thought up the clever name "manifold-with-boundary". But lots of things were already known about manifolds that were not true for these new things. (For instance: manifolds have no boundary!) So everyone agreed that m-with-b was a new category, one that included (but was not included in) the old category of "just plain old manifolds," because otherwise they'd have to go rewrite everything from the past replacing "manifold" with "manifold-without-boundary". Ugh!

The choice to include the sphere, say, in the collection of manifolds-with-boundary, even though its boundary is empty, might seem peculiar. But it didn't take long for folks to see that if they DIDN'T do this, then lots of theorems would start out "Let M be a manifold or manifold-with-boundary", i.e., lots and lots of the things proved for manifolds-with-nonempty-boundary turned out to also be true for manifolds with empty boundary (i.e., ordinary manifolds). So they made the new term inclusive rather than exclusive.

Of course, this little bit of history is mostly fiction --- there wasn't a congress of mathematicians who all agreed that henceforth they'd do things THIS way. It took a while to settle out. And in the intervening years, I'm sure there have been folks who decided that the old way was stupid, and they were going to change the world and insist on manifold-without-boundary and manifold-with-boundary, and let the word "manifold" die a well-earned death. I don't believe that's caught on, but I could be wrong.

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    $\begingroup$ In terms of the whole "manifold-with-boundary is not a manifold" discussion. As you say, in english, an "adjective noun" is a special kind of "noun" while in this case, it's not true. On the other hand, this grammatical construction does seem to be fairly common in mathematics. Compare "left inverse" with "inverse", "left ideal" with "ideal", "skew field" with "field", "rational sphere" with "sphere", etc. $\endgroup$ Aug 31, 2020 at 15:43
  • $\begingroup$ Ooh...nice point, Jason. I think that it might be nice for textbook authors, at least at the "introducing manifolds" or "introducing abstract algebra" level, to mention this habit among mathematicians. I don't recall seeing this in Linear Algebra, although I suppose that "left-inverse" might show up in some texts. The more we can explain/clarify our terminology, the better. I don't mean to say students shouldn't be confused at time, but they should be confused about the right things, not "how come an X-with-Y isn't also an X?" kinds of things. :) $\endgroup$ Sep 1, 2020 at 15:40
  • $\begingroup$ @JohnHughes - Thanks. I'm going to have to take some time to digest that. Up until now I just assumed some manifolds had boundaries and some didn't - end of story. I've just discovered that Lee goes into some detail on this in the first chapter of his "Introduction to Smooth Manifolds". Don't suppose you have a reference for the "pullbacks via homotopic maps give the same integral" theorem? I've looked through Lee, Spivak, Munkres and others but can't find it anywhere (which doesn't mean it's not there, just that I couldn't find it). $\endgroup$
    – Peter4075
    Sep 1, 2020 at 17:26
  • $\begingroup$ It's exactly the theorem that you're talking about --- the one that says $$\intop_{M}f^{*}\omega=\intop_{M}g^{*}\omega.$$ whenever $f,g$ are homotopic maps from a manifold $M^n$ to a manifold $K$, and $\omega$ is a closed $n$-form on $K$ (should have mentioned closed-ness in my too-glib shorthand). It's pretty much the starting point for deRham cohomology theory. $\endgroup$ Sep 1, 2020 at 17:36
  • $\begingroup$ @JohnHughes: I agree. I wasn't even consciously aware of this grammatical flip until I was teaching an "Intro to proofs" class. In the context of function composition, one has a notion of left and right inverses. I was confused by several students' wrong proofs until it finally clicked - "Oh, you're assuming a left inverse is an inverse!" $\endgroup$ Sep 1, 2020 at 17:36

Summary (Based on comments): The equality $$\int f^*\omega=\int g^*\omega$$ does not hold in general (e.g. manifolds with boundary). Actually as I searched, manifold with boundary has been defined on page 380 and after that he considered manifold and manifold with boundary distinctive but I couldn't find any remark in the book about avoiding any possibility of ambiguity of these terms. As Shifrin said, the author use redundant term "with boundary" to refer to a manifold with boundary. i.e. he explicitly says with boundary if boundary is allowed.

Here is also a related post.


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