Homotopy equivalence upper triangular matrices and torus In an old algebraic topology exam, I came across this question.

Let $G$ be the set of invertible upper triangular matrices in $\mathbf{C}^{2\times 2}$, as a topological subspace of $\mathbf{C}^3\cong \mathbf{R}^6$.
(a) Prove that $G$ is homotopy equivalent to the torus.
(b) Determine the push forward $\det_*:\pi_1(G,\mathbf{1})\longrightarrow \pi_1(\mathbf{C}^*,1)$ of the determinant map $\det:G\longrightarrow \mathbf{C}^*:A\longmapsto \det A$.

(a) Since $G$ consists of all matrices $\begin{bmatrix}a&b\\0 & c\end{bmatrix}\in\mathbf{C}^{2\times 2}$ with $a,c\neq 0$, it can be identified with $\mathbf{C}^*\times\mathbf{C}\times\mathbf{C}^*$. This can be used to prove that $G$ is path connected.
Identifying the torus $T$ with $S^1\times S^1$, we need to find continuous $f:G\to T,g:T\to G$ such that $f\circ g\simeq\text{id}_T$ and $g\circ f\simeq \text{id}_X$.
We let
$$f\left( \begin{bmatrix}a&b\\0&c\end{bmatrix}\right)=(\operatorname{arg}a,\operatorname{arg}b),\quad g(\theta,\phi)=\begin{bmatrix}\cos\theta+i\sin\theta & 1 \\ 0 & \cos\phi+i\sin\phi \end{bmatrix}.$$
where $\operatorname{arg}$ sends a complex number to its argument in $[0,2\pi)$.
Then $f\circ g=\text{id}_T$ and $g\circ f:G\to G:\begin{bmatrix}a&b \\ 0 & c\end{bmatrix}\mapsto \begin{bmatrix} a/||a|| & b/||b|| \\ 0 & c/||c|| . \end{bmatrix}$. Since $G$ is path connected, $g\circ f\simeq \text{id}_G$.
Is this correct?
(b) I know that the fundamental group of the product space $\mathbf{C}^*\times\mathbf{C}\times\mathbf{C}^*$ is the direct product of the fundamental groups, thus $\pi_1(G,\mathbf{1})=\mathbf{Z}\times\mathbf{Z}$. (I think the homotopy equivalence cannot be showed by simply noting that the fundamental group of the torus is also $\mathbf{Z}\times\mathbf{Z}$.) This has two generators $(1,0)$ and $(0,1)$. We can represent (is this correct?) these on the level of the fundamental group by the loops
$$\gamma_0:I\to G:t\mapsto \begin{bmatrix}e^{it} & 0 \\ 0 & 1\end{bmatrix},\quad \gamma_1:I\to G:s\mapsto \begin{bmatrix}1 & 0 \\ 0 & e^{is}\end{bmatrix},$$
whose images under $\det_*$ are respectievely $I\to \mathbf{C}^*:t\mapsto e^{it}$ and $I\to \mathbf{C}^*:s\mapsto e^{is}$. This means that the push forward morphism is given by $\mathbf{Z}\times\mathbf{Z}\longrightarrow \mathbf{Z}:\begin{cases}(1,0)\longmapsto 1 \\ (0,1) \longmapsto 1 \end{cases}$.
Is this correct?
 A: Your idea for (a) is nice, but how can be sure that your maps $f, g$ are continuous? Here is an alternative definition:
$$f\left( \begin{bmatrix}a&b\\0&c\end{bmatrix}\right)=(a/\lvert a \rvert,c/\lvert c \rvert) \quad g(v,w)=\begin{bmatrix}v & 0 \\ 0 & w\end{bmatrix}.$$
Then $f \circ g = id_T$ and
$$(g \circ f) \left( \begin{bmatrix}a&b\\0&c\end{bmatrix}\right)=\begin{bmatrix}a/\lvert a \rvert&0\\0& c/\lvert c \rvert\end{bmatrix} .$$
A homotopy $H :g \circ f \simeq id_G$ is defined by
$$H(\left( \begin{bmatrix}a&b\\0&c\end{bmatrix}, t \right) =  \begin{bmatrix}(1-t)a/\lvert a \rvert + ta & tb\\0& (1-t)c/\lvert c \rvert + tc \end{bmatrix} .$$
For (b) consider the map $h = det \circ g : T \to \mathbb C^*$. We have $h(v,w) = vw$. $\pi_1(T,(1,1))$ has two generators given by $\gamma_0(t) = (e^{2\pi it},1)$ and $\gamma_1(t) = (1,e^{2\pi it})$. Hence $(h \circ \gamma_i)(t) = e^{2\pi it}$ which is the generator of $\pi_1(\mathbb C^*,1)$.
Thus your answer for (b) is correct.
A: For (a) one can pretty explicitly see that it is homotopy equivalent to the torus just by manipulating the spaces involved. Note that your space of matrices is homeomorphic to the space $\left\{\begin{bmatrix}a&0\\0 & c\end{bmatrix}\bigg| \mathbb{C} \ni a,c \neq 0 \right\} \times \mathbb{C}$. This is homotopy equivalent to $\left\{\begin{bmatrix}a&0\\0 & c\end{bmatrix}\bigg| \mathbb{C} \ni a,c \neq 0 \right\}$ since $\mathbb{C}$ is contractible, and $\left\{\begin{bmatrix}a&0\\0 & c\end{bmatrix}| \mathbb{C} \ni a,c \neq 0 \right\}$ is homeomorphic to $\mathbb{C}^\times \times \mathbb{C}^\times$ which is homotopy equivalent to $S^1 \times S^1$.
