Proving that $\mathbb{R}/\mathbb{Z} \cong S^1$ as groups I am trying to prove that $\mathbb{R}/\mathbb{Z}$ under addition is isomorphic to $S^1$. The definitions I have to work with are:
\begin{align*} 
\mathbb{R}/\mathbb{Z} & = [0,1) \text{ with addition mod $1$} \\ 
S^1 & =  \{z \in \mathbb{C}^{*} \mid |z| = 1\} \text{ with multiplication.} 
\end{align*}
Though it seems natural to use the first isomorphism theorem, I don't yet have access to it, so I need to use the bijection $t \mapsto \exp(2 \pi i t)$. Here is my attempt at a proof.

Define $\varphi: \mathbb{R}/\mathbb{Z} \to S^1$ by $\varphi(t) = \exp(2 \pi i t)$. We first show that $\varphi$ is a homomorphism.
Homomorphism. Let $a,b \in \mathbb{R}/\mathbb{Z}$. We denote addition in $\mathbb{R}/\mathbb{Z}$ by $*$ and addition in $\mathbb{R}$ by $+$. If $a + b < 1$, then $a * b = a + b$, and we have
\begin{align*}
\varphi(a * b) & = \varphi(a + b) \\
& = \exp^{2 \pi (a + b) i} \\
& = \exp(2 a \pi i + 2 b \pi i) \\
& = \exp(2 a \pi i) \exp(2 b \pi i) \\
& = \varphi(a) \varphi(b).
\end{align*}
Now suppose that $a + b \geq 1$. Then $a*b = a + b - 1$, and we have
\begin{align*}
\varphi(a * b) & = \varphi(a + b - 1) \\
& = \exp(2 \pi (a + b - 1) i) \\
& = \exp(2 \pi a i + 2 \pi b i - 2 \pi i) \\
& = \exp(2 \pi a i) \exp(2 \pi b i) \exp(-2 \pi i) \\
& = \varphi(a) \cdot \varphi(b) \cdot 1 \\
& = \varphi(a) \varphi(b), 
\end{align*}
so $\varphi$ is a bijective homomorphism and therefore an isomorphism.

We next show that $\varphi$ is a bijection of sets.

Injectivity.
For injectivity, we show that $\mathrm{ker}(\varphi) = \{0\}$. For any $t \in [0,1)$, we have
\begin{align*}
t \in \mathrm{ker}(\varphi) & \iff \varphi(t) = \exp(2 \pi i t) = 1 \\
& \iff \cos(2\pi) + i \sin(2\pi t) = 1 + 0i \\
& \iff \sin(2 \pi t) = 0 \\
& \iff t \in \mathbb{Z} \cap [0,1) \\
& \iff t = 0,
\end{align*}
so $\mathrm{ker}(\varphi) = \{0\}$, so $\varphi$ is injective.
Surjective. Let $z \in S^1$, and write $z$ in its exponential form $z = \exp^{i \theta}$ where $\theta \in [0,2\pi)$. As $0 \leq \theta < 2\pi$, we have $0 \leq \frac{1}{2\pi} \theta < 1$, so $\frac{1}{2\pi} \theta \in \mathbb{R}/\mathbb{Z}$. We then have
$$
\varphi\left(\frac{1}{2\pi} \theta\right) = \exp\left(2\pi \cdot \frac{\theta}{2\pi} i \right) = \exp(i \theta) = z,
$$
so $\varphi$ is surjective and therefore bijective.

 A: Well it's true that you cannot "use" the first isomorphism theorem. But there is no harm in proceeding like the proof of the first isomorphism theorem.
Define $q:\Bbb{R\to R/Z}$ by $q(x) = (x-\lfloor x\rfloor) +\Bbb{Z}$ . This is just the quotient map which is sending a real number to it's coset representative.
This map is clearly surjective as for any $y+\Bbb{Z}$ such that $y\in[0,1)$ we can take the pre-image to be $y\in\Bbb{R}$.
Define $f:\mathbb{R}\to S^{1}$ by $f(x)=e^{2i\pi x}$ .
Then it is easy to verify that this is a homomorphism.
Define the map $\bar{f}:\Bbb{R}/\Bbb{Z}\to S^{1}$ such that $\bar{f}(x+\Bbb{Z})=f(x)=e^{2i\pi x}$.
Now we need to see if this map is well-defined.
if $x+\Bbb{Z}=y+\Bbb{Z}$. Then $x-y\in\Bbb{Z}$ so let $x=y+n$ for some $n\in\Bbb{Z}$.
Then we have $\bar{f}(x+\Bbb{Z})=\bar{f}((y+n)\Bbb{Z})=e^{2i\pi (y+n)}=e^{2i\pi y}\cdot e^{2i\pi n} =e^{2i\pi y}=\bar{f}(y+\Bbb{Z})$.
So this is well defined.
Now if $x+\Bbb{Z},y+\Bbb{Z} \in \Bbb{R}/\Bbb{Z}$ then
$$\bar{f}((x+\Bbb{Z})+(y+\Bbb{Z}))=\bar{f}(x+y+\Bbb{Z})=f(x+y)=e^{2i\pi(x+y)}$$
$$=e^{2i\pi x}\cdot e^{2i\pi y}=\bar{f}(x+\Bbb{Z})\cdot \bar{f}(y+\Bbb{Z})$$
and hence $\bar{f}$ is a group homomorphism.
Now you have already correctly verified that $\ker{\bar{f}}=\Bbb{Z}$ which is the identity in $\Bbb{R}/\Bbb{Z}$. and it is surjective as $f$ and $q$ are surjective. In otherwords you can find $\frac{\theta}{2\pi}\in \Bbb{R}$ such that $\bar{f}(\theta+\Bbb{Z})=f(\theta)=e^{i\theta}\in S^{1}$.
Hence this is an isomorphism.
Note that we are implicitly using the Universal Propert of Quotients. That is $q$ and $f$ uniquely yields a homomorphism $\bar{f}$ such that $\bar{f}\circ q=f$. In fact you need not even verify the surjectivity of $\bar{f}$ as the surjectivity of $f$ and $q$ guarantee it due to the universal property of quotients.
