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Let $f : \mathbb{S}^n \to \mathbb{S}^n$ be a continuous function.

From homology, the degree $\deg_1(f)$ of $f$ can be defined as the integer $n$ such that $f_* : x \mapsto n \cdot x$, where $f$ induces the morphism $f_* : \mathbb{Z} \simeq H_n(\mathbb{S}^n) \to \mathbb{Z} \simeq H_n(\mathbb{S}^n)$.

From topological degree, the degree $\deg_2(f)$ of $f$ can be defined as $\deg(\tilde{f},B^{n+1},0)$, where $\deg$ is the Brouwer degree as defined in the book Topological Degree Theory and Applications or in the paper An Elementary Analytic Theory of the Degree of Mapping in n-Dimensional Space and $\tilde{f} : \overline{B}^{n+1} \to \mathbb{R}^{n+1}$ is a continuous extension of $f$ on the unit closed ball of $\mathbb{R}^{n+1}$.

(The integer $\deg_2(f)$ does not depend on the choice of $\tilde{f}$.)

Question: Do $\deg_1$ and $\deg_2$ coincide?

I think it is true, but probably I do not have enough knowledge about homology theory to find a rigorous proof...


Added: Let $\Omega \subset \mathbb{R}^n$ be a bounded open set, $f : \overline{\Omega} \to \mathbb{R}^n$ be a continuous function $C^1$ on $\Omega$ and $p \notin f(\partial \Omega)$ a regular value of $f$. The Brouwer degree of $f$ is defined by $$\deg(f,\Omega,p)= \sum\limits_{x \in f^{-1}(p)} \mathrm{sign}(J_f(x))$$ where $J_f$ is the Jacobian determinant of $f$.

It can be shown that $\deg(g,\Omega,q)=\deg(f,\Omega,p)$ if $q$ is a regular value of a function $g \in C^1(\overline{\Omega})$ such that $\sup\limits_{x \in \overline{\Omega}} \|f(x)-g(x)\|$ and $\|p-q\|$ are sufficiently small.

Using Sard theorem and Stone-Weierstrass approximation theorem, we can define the Brouwer degree of a continuous function $f: \overline{\Omega} \to \mathbb{R}^n$ with respect to a bounded open set $\Omega \subset \mathbb{R}^n$ and a point $p \notin f(\partial \Omega)$ by $$ \deg(f,\Omega,p)=\lim\limits_{n \to + \infty} \deg(f_n,\Omega,p_n)$$ where $(f_n)$ is a sequence in $C^1(\overline{\Omega})$ converging uniformly to $f$ and $p_n$ is a regular value of $f_n$ for each $n$.

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  • $\begingroup$ I don't know the definition of Brouwer degree... But every continuous map from cpt mfd to mfd can be homotopic to a smooth one, does that help? (I assume the definition of Brouwer's doesn't change when you homotopy the map) $\endgroup$ – lee Jul 4 '13 at 16:14
  • $\begingroup$ @DanielRust: A rigorous definition is rather long to state. However, I added some details. $\endgroup$ – Seirios Jul 4 '13 at 16:19
  • $\begingroup$ @lee: Maybe, I will think about that, thank you. By the way, do you have a reference for this result? $\endgroup$ – Seirios Jul 4 '13 at 16:20
  • $\begingroup$ @Seiros Thanks for adding the extra detail. $\endgroup$ – Dan Rust Jul 4 '13 at 16:27
  • $\begingroup$ Which result? The approaximation by smooth map? See the 'Introduction to Smooth Manifolds' written by John. Lee, Chap 10 if my memory doesn't go wrong. $\endgroup$ – lee Jul 5 '13 at 3:31
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I think the answer to your question is yes. Here is the sketch of proof. First we may assume this map is smooth. By Sard's thm, we choose a noncritical value, say, p. Then p must have preimages $x_i$, i=1,...,k. Now the definition of Brouwer's degree is the sum of 1 or -1 for each $x_i$ depending on the Jacobi is more(1) or less(-1) than 0. Now if you can see page 136 of Hatcher's Algebraic topology. You'll find the definition of local degree of f at $x_i$, you can check easily that the local degree is 1 if the map is locally orientation preserving, i.e. with positive Jacobi and vice versa. And the homological definition of degree equals the sum of local degrees, see Proposition 2.30 of Hatcher's algebraic topology. I think I just give a proof, though some details are left to be checked.

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You may find the answer looking to Amann-Weiss theorem (1972) which illustrates the equivalence of different degree theories. There was published in the article 'On the uniqueness of topological degree' by Amann.

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