Identify the points $0$ and $1$ on $[0,1]$ and show that the resulting space $[0,1]/\sim$ is homeomorphic to $S^1$.
I'm trying to figure out what the conditions are for a map $[0,1] \to S^1$ to induce a homeomorphism. If $f : [0,1] \to S^1, t \mapsto e^{2\pi i t}$, then $f$ is continuous and surjective and we now have that if $f$ is a quotient map, then it induces an homeomorphism $\tilde{f} : [0,1]/\sim \to S^1$.
How can we verify that $f$ is a quotient map? One way is to show that $V$ is open in $S^1$ iff $f^{-1}(V)$ is open in $[0,1]$. Another way would be to show that $f$ has a right inverse?
What I'm wondering is that is it enough to conclude that $f(0) = f(1)$ in this case?
In total generality, the most basic method is to show that the continuous surjection $f:X\to Y$ is also a quotient map, meaning that sets in $Y$ are open iff. their preimage is open. Equivalently, you check that $f$ is carries open saturated sets to open sets. Under an equivalence relation $\sim$ on $X$, we say a set $A\subseteq X$ is $(\sim)$-saturated if: $$x\in A\implies[x]_\sim\subseteq A$$
Here, $\sim$ is taken to be the relation: $$x=y\iff f(x)=f(y)$$I.e. the relation of the fibres of $f$. This allows you to think about the quotient topology without worrying about the relations explicitly. My comfortability with quotient spaces was increased by reading the relevant sections of "Topology and groupoids" by Ronnie Brown.
However, when dealing with compact spaces the situation is much easier to think about. If $X$ is compact and $Y$ is Hausdorff, a continuous surjection $f:X\to Y$ is automatically a quotient map. For, any open set of $X$ is the complement of a compact set hence the $f$-image is again the complement of a compact set. As $Y$ is compact, this makes the $f$-image open. So $f$ is in fact an open map, in this case, which is a little stronger than required (being open on the open saturated sets).
Since $t\mapsto e^{2\pi it}$ is a continuous surjection $[0,1]\to S^1$, with the LHS compact and the RHS Hausdorff, this is necessarily a quotient map. A quotient by what relation? Well, the induced relation is precisely that of $0\sim1$, as this map has fibres either singletons or $\{0,1\}$. In other words, $e^{2\pi it}=e^{2\pi is}$ iff. $s=t$ or $\{s,t\}\subseteq\{0,1\}$.
N.B. I often find it easier to think about quotients in the following categorical way: $$f:X\to Y$$Is a quotient map iff. for all continuous $g:X\to Z$ with $g$ constant on every $f$-fibre, (that is, $f(x)=f(y)\implies g(x)=g(y)$) then $g=hf$ for a unique $h:Y\to Z$ ($g$ 'factors through' $Y$).