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This is a question from a qualifying exam. Let $X$ be a compact, oriented $n$-dimensional manifold. Show that for any $k \in \mathbb{Z}$, there exists a continuous map $f: X \to S^n$ of degree $k$.

I was pleased with the very nice solution via suspensions at For every $k \in {\mathbb Z}$ construct a continuous map $f: S^n \to S^n$ with $\deg(f) = k$. in the case where $X = S^n$. It seems that for this question, it should suffice to show that there exists a degree $\pm 1$ map from $X$ to $S^n$, and then we can compose with a degree $\pm k$ self-map of $S^n$ to get a degree $k$ map from $X$ to $S^n$ (because the degree of a composition of maps is the product of the degrees of the component maps).

One idea I've had so far is to consider an embedding $X \to \mathbb{R}^N \backslash \{0\}$ for some large $N$ and then project onto the $n$-sphere, but I do not know how one would guarantee that this would have degree 1. And while this isn't necessary to answer the question, what would a degree 1 map $T^2 \to S^2$ even look like? I cannot easily visualize such a map.

EDIT: To answer my own question about a degree 1 map $T^2 \to S^2$ that is slightly less `singular' than Jared's answer below: imagining the sphere and torus as their nice looking ball and doughnut shapes, just place the sphere inside of the torus (i.e. in the tube itself, not at its center of mass) and then project the the torus onto the sphere.

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    $\begingroup$ You can use the fact (in Hatcher) that the sum of local degrees is equal to the degree. Locally each induced map on homology is a map from spheres (of 1 lower dimension) and you know how to get any degree you want! $\endgroup$ – Elden Elmanto Jul 26 '13 at 20:38
  • $\begingroup$ In fact much stronger statement is true: (in the assumptions of the question) $\operatorname{deg}\colon[X,S^n]\to\mathbb Z$ is an isomorphism (Hopf theorem). $\endgroup$ – Grigory M Jan 17 '14 at 20:36
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Here is an answer that uses more abstract machinery. Let $B \subset X$ be an open set homeomorphic to an open ball in $\Bbb R^n$. Let $p : X \to X / (X - B)$ be the quotient map. We have $X/(X - B) = S^n$. By the naturality of the long exact sequence, we have the commutative diagram: $$ \require{AMScd} \begin{CD} H_n(X) @>\cong>> H_n(X, X - B)\\ @VVq_*V @VV\cong V \\ H_n(X / (X - B)) @>\cong>> H_n(X / (X - B), (X - B) / (X - B)) \end{CD} $$

The upper map is an isomorphism since $X$ is orientable. The right map is an isomorphism by excision (see Hatcher's Algebraic Topology, 2.22). The lower map is an isomorphism since $(X - B) / (X - B)$ is a single point.

It follows that $q_*$ is an isomorphism. Hence $q$ has degree one as desired.

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    $\begingroup$ Wow! I love it! $\endgroup$ – Jon Paprocki Dec 27 '13 at 5:25
  • $\begingroup$ Excuse me, but why is $X/(X-B)=S^n$, and how does orientability of $X$ imply $H_n(X)\cong H_n(X,X-B)$? (Is that from Hatcher? I could not find it.) $\endgroup$ – Clara Sep 11 '14 at 7:12
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    $\begingroup$ @Clara 1) It is equivalent to $D^n / \partial D^n = S^n$. 2) See theorem 3.26.a. $\endgroup$ – Ayman Hourieh Sep 11 '14 at 12:42
  • $\begingroup$ Can you explain why $q_{*}$ being an isomorphism implies that $q$ is a degree one map? $\endgroup$ – Michael Aug 14 '16 at 23:46
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    $\begingroup$ @Michael An isomorphism $\mathbb{Z} \to \mathbb{Z}$ must map $1$ to $\pm 1$. Otherwise it would have no hope of being surjective. If it maps to $-1$, you can just replace $q_*$ by $-q_*$ to get degree $1$. $\endgroup$ – Anschel Schaffer-Cohen Sep 11 '17 at 2:02
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Here is the (rough) idea for constructing a map from $X$ to $S^n$ of degree $1$. Take any point $p\in X$ and map it to a pole of $S^n$. Next, smoothly map a small open ball around $p$ to the rest of $S^n$ minus the opposite pole. Then map every other point in $X$ not in this open ball to the opposite pole.

In essence, you're wrapping the sphere with a disc inside of $X$, and bunching everything else into a point.

This map has degree one because it is a local, orientation-preserving diffeomorphism around $p$.

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  • $\begingroup$ Ah! I was expecting some abstract machinery rather than a constructive answer, so this is very nice. I want to hold off on accepting an answer just yet, to see if anybody has a homological argument or something. $\endgroup$ – Jon Paprocki Jul 26 '13 at 17:31
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    $\begingroup$ In fact this answer easily shows hot to get a map of degree $k$. Do the same procedure for balls around $k$ points of $X$. $\endgroup$ – Cheerful Parsnip Dec 27 '13 at 1:30

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