For every $k \in {\mathbb Z}$ construct a continuous map $f: S^n \to S^n$ with $\deg(f) = k$. Suppose $S^n$ is an $n$-dimensional sphere.
Definition of the degree of a map:
Let $f:S^n \to S^n$ be a continuous map. Then $f$ induces a homomorphism $f_{*}:H_n(S^n) \to H_n(S^n)$ . Considering the fact that $H_n(S^n) = \mathbb {Z}$ , we see that $f_*$ must be of the form $f_*(n)=an$ for some fixed integer $a$. This $a$ is then called the degree of $f$.
Question: For every $k \in {\mathbb Z}$ how does one construct a continuous map $f: S^n \to S^n$ with $\deg(f) = k$?
 A: Here is another solution.
Claim 1: If $f:S^n \to S^n$ has degree $d$ then so does $\Sigma f: S^{n+1} \to S^{n+1}$
Proof: Use the Mayer-Vietrois sequence for $S^{n+1}$. Let $A$ be the complement of the North pole, and $B$ the complement of the South pole. Then $S^n \simeq A \cap B$ and the connecting map $\partial_*$ in the Mayer-Vietrois sequence is an isomorphism. We get the following commutative diagram
$$
\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!}
\newcommand{\la}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xleftarrow{\quad#1\quad}\!\!\!\!\!\!\!\!}
\newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.}
%
\begin{array}{llllllllllll}
H_{n+1}(S^{n+1}) & \ra{\partial_*} & H_n\left(A \cap B\right) & \la{i_*} & H_n(S^n)\\
\da{\Sigma f_*} & & \da{} & & \da{f_*} \\
H_{n+1}(S^{n+1}) & \ra{\partial_*} & H_n\left(A \cap B\right) & \la{i_*} & H_n(S^n)\\
\end{array}
$$
in which each horizontal map is an isomorphism. Thus $\Sigma f_* = \partial_*^{-1} i_* f_* i_*^{-1}\partial_*$ and applying homology shows that $\text{deg}(f) = \text{deg}(\Sigma f)$
Thus we are reduced to simply showing that there is a map $f:S^1 \to S^1$ of degree $k$. But this is just the winding number and it is (reasonably well known) that the map $z \mapsto z^k$ (where we view $S^1$ as the unit circle in $\mathbb{C}$) has degree $k$. 
Finally I would direct you to have a look at Algebraic Topology by Hatcher:


*

*Example 2.31 gives a direct construction of a map of arbitrary degree; 

*Example 2.32 works through the calculation of the map $f(z)=z^k$ proving it has degree $k$; and

*Prop 2.33 gives another prove of Claim 1 above (which basically takes a different route to the same commutative diagram).

A: Hint: Parameterize $S^n$ as 
$$g(t_1,\ldots,t_n)=\begin{pmatrix}
\cos(t_1)\\
\sin(t_1)\cos(t_2)\\
\vdots\\
\sin(t_1)\cdots\sin(t_{n-1})\cos(t_n)\\
\sin(t_1)\cdots\sin(t_{n})
\end{pmatrix}$$
over $t_1,\ldots,t_{n-1}\in [0,\pi]$ and $t_n\in [0,2\pi)$, and define $f:S^n\to S^n$ by $$f(g(t_1,\ldots,t_{n-1},t_n))=g(t_1,\ldots,t_{n-1},kt_n)$$
which amounts to wrapping the sphere around itself $k$ times in the $x_nx_{n+1}$ plane. One way to show that this is the desired map is to show that it is continuous and then consider one of the simplices perpendiclar to the $x_nx_{n+1}$ plane, when you view the simplicial complex of $S^n$ as a subdivided version of $\partial I^{n+1}$ sitting in $E^{n+1}$ (for example, the top edge of a square, top face of a cube, etc).
