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 $\deg(f) = k$?

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).
