The $n$-disk $D^n$ quotiented by its boundary $S^{n-1}$ gives $S^n$ Define $D^n = \{ x \in \mathbb{R}^n : |x| \leq 1 \}$. By identifying all the points of $S^{n-1}$ we get a topological space which is intuitively homeomorphic to $S^n$.
If $n = 2$, this can be visualised by pushing the centre of the disc $D^2$ down so you have a sack, then shrinking the boundry of the sack to a point which gives you a teardrop shaped object which is clearly homeomorphic to $S^2$.

I am new to algebraic topology. How do I prove that the quotient space is actually homeomorphic to $S^n$. I haven't been able to write down explicitly a continuous map between $D^n$ and $S^n$ which maps $S^{n-1}$ to a point on $S^n$, which at the moment is the only way I know how to begin showin that two spaces are homeomorphic.
Is more machinery needed? If so I am interested to hear what is needed. If not, please tell me how stupid I am and give me a hint!
None of the answer give any map explicitely so I'll write one here. Note that $S^{n−1}\times I→D^n,(u,t)\mapsto tu$ (this is scalar product) is a quotient map and thus one can define (after verifying some things) the continuous map $D^n\to S^n,tu\mapsto (\sin⁡(tπ)u,\cos⁡(tπ))$ which can be verified to work.
You know that $S^n$ is a one point compactification of $R^n$. That is $S^n = R^n \cup \{\infty\}$ with appropriate topology. Now define a map $f\colon D^n \to S^n$ as follows if $x$ is in interior of $D^n$ then $f(x) = \frac{x}{1 -\mod{x}}$ but if $x$ is in boundary then $f(x) = \infty$. Now it is easy to see that $f$ is continuas in the interior and as well as in boundary. Sequential criterion may help you to conclude the statement. Now $D^n$ is compact so the map is a closed map and therefore a quotient map. Hence it induces a homeomorphism from $D^n$ to $S^n$.