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I've been reading some articles/books on degree theory and I've noticed a detail that I can't quite understand fully yet.

Some articles define the degree of a smooth function $f:M \to N$ (where $M$ and $N$ are supposed to be compact, connected, oriented, smooth manifolds of the same dimension $n$) as $$ \text{deg}(f) = \sum_{x \in f^{-1}(q)} \text{sign}_x(f), $$ where $q \in N$ is a regular value of $f$ (that is, every point on its preimage satisfies that the differential on that point is onto. One can see that under these assumptions on the manifolds $M$ and $N$ this set $f^{-1}(q)$ is finite and thus the sum is well-defined), and where $\text{sign}_x(f)$ equals $1$ if $df_x$ preserves orientation, and $-1$ if it reverses it (I believe one can see that an oriented manifold has only two possible orientations), in the sense that $$ \text{sign}_x(f) = \frac{\det(J_f(x))}{|\det(J_f(x))|}, $$ where $J_f$ denotes the jacobian matrix of the coordinate representation of $f$.

The thing is, I've seen other articles where the assumptions on $M$ and $N$ are supposedly weaker. That is, they suppose they are oriented, smooth and $\textbf{without boundary}$ (I assume connectedness is implied as well but I'm not 100% sure about this).

So, my question is, what's really the difference between assuming compactness and empty boundary and why is the second one weaker? Is there any other assumption that has to be made in this case? Why is the condition that the manifolds have no boundary important when defining the degree? I have the intuition (given some things I've read) that maybe it's because if the manifold has boundary, then the invariance under homotopy would break, or something of the likes.

If someone has any good and intuitive explanation then I would really appreciate it. Also, I'm barely starting on this topic so maybe some of the things I've presented on this post are wrong or unclear. If so, please let me know as well. Thanks!

Edit: as someone in the comments rightfully said, I forgot to cite my sources. There are two specifically that I'm referring to:

  1. Polo Gómez, J. (2020). Brouwer-Kronecker degree theory. Universidad Complutense de Madrid. This article is where degree is defined using manifolds without boundary. The work is in Spanish but I believe it is very well crafted.

  2. Lee, J. M. (2013). Introduction to smooth manifolds (2.ª ed.). Springer. https://doi.org/10.1007/978-1-4419-9982-5. In section 17 Lee shows a brief introduction to in degree theory (in relation to the integral of $n$-forms and De Rham cohomology), and defines the degree using compact manifolds.

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    $\begingroup$ Which articles? Always cite your sources! $\endgroup$ Commented yesterday
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    $\begingroup$ For many authors, manifolds, by default, do not have boundary. Check the terminology used by your sources. $\endgroup$ Commented yesterday
  • $\begingroup$ Incidentally, there is a version the degree theory for manifolds with boundary and for noncompact manifolds, but it is more subtle. For compact manifolds with boundary, one works with maps of pairs $f: (M,\partial M)\to (N,\partial N)$ and the corresponding notion of homotopy of maps. $\endgroup$ Commented yesterday
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    $\begingroup$ Standard sources, other than Lee, are Guillemin/Pollack and Hirsch’s Differential Topology. $\endgroup$ Commented yesterday
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    $\begingroup$ Sidenote on the connectedness. A manifold has only finitely many connected components. So in general assuming a manifold is connected is sort of without loss of generality. One can apply the theorem to each connected component and then 'add together' the result. What 'add together' means depends on the specific theorem but no fundamentally new or surprising things happen if you have several connected components. $\endgroup$ Commented yesterday

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To give the question a proper answer: In his book (Chapter 17, Theorem 17.35 and following discussion), Lee assumes that manifolds in question have empty boundary (this is his definition of manifolds in chapter 1 of the book). Without this assumption, Theorems 17.35 and 17.36(c) are simply false.

Examples are quite easy. For instance, you can take the identity map $f: [0,1]\to [0,1]$ and the constant map $g: [0,1]\to [0,1]$, trivially homotopic to $f$. Then $\deg (f)=1$ while $\deg(g)=0$. Similarly, if one takes the map $h: [0,1]\to [0,1]$, $h(x)=x/2$, then the degree of $h$ is not even well-defined since it depends on the choice of a point $y\in [0,1]$. Also, $\int_{[0,1]} dx=1$, while $\int_{[0,1]} h^*(dx)=1/2$.

There are (very) useful notions of degree for maps between noncompact manifolds and manifolds with boundary. For noncompact manifolds (without boundary), one has to restrict to proper maps, proper homotopies and compactly supported cohomology groups. Similarly, for compact manifolds with boundary $M, N$, one has to restrict to maps of pairs $(M, \partial M)\to (N, \partial N)$ and homotopies of maps of pairs. For de Rham cohomology, one has to use the relative cohomology groups $H^n(M, \partial M)$. Then everything goes through as in the setting of compact manifolds without boundary.

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  • $\begingroup$ Thanks for the answer and the remarks! I still don't get why assuming empty boundary in the cases Lee studies is completely necessary. Could you give me a brief explanation on why the manifolds have to have empty boundary in order for them to work? $\endgroup$ Commented yesterday
  • $\begingroup$ @Pabloo: Try to construct counter-examples to the theorems I mention in the case when $M$ is a closed interval. If you understand the material, it should take you less than an hour. If you cannot do it in an hour, let me know. $\endgroup$ Commented yesterday
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    $\begingroup$ It's not a mistake -- it's just a terminological convention. Notice that I explained in Chapter 1 of ISM (see the bottom of page 26) that the term "manifold" without further qualification means manifold without boundary. $\endgroup$ Commented yesterday
  • $\begingroup$ @JackLee: Oh, OK, this does clarifies the issue. The trouble is that many, if not most, people do not read books "linearly," hence, a convention in chapter 1 goes unnoticed when reading chapter 17. $\endgroup$ Commented yesterday
  • $\begingroup$ If you take $M=N=[0,1]$, and then $f_0(x) = x$ the identity map and $f_1(x)=\frac{1}{2}$ a constant mapping (both from $[0,1]$ to $[0,1]$). Then $\text{deg}(f_0)=1$ and $\text{deg}(f_1)=0$, but these maps are homotopic since the homotopy $H(x,t)=(1-t)x + t \cdot \frac{1}{2}$ takes one to the other. Is this example correct? Maybe it is too simple. $\endgroup$ Commented yesterday

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