Set $\mathcal{F}:=\{ A \in \text{SL}\_2(\mathbb{R}) \, | \, Ae\_1 \in \operatorname{span}(e\_1) \, \, \text{ and } \, \, A \, \text{ is not conformal} \,\}$, and

$\mathcal{NC}:=\{ A \in M\_2(\mathbb{R}) \, | \det A \ge 0 \, \,\text{ and } \, A \text{ is not conformal} \,\}$.

By a non-conformal matrix, I mean a matrix whose singular values are distinct. (i.e. I allow non-zero singular matrices in $\mathcal{NC}$).

> 
> Is each connected component of $\mathcal{F}$ contractible in $\mathcal{NC}$?
> 
> 
> 

$\mathcal{F}$ has two connected components, both homeomorphic to an open half-plane with one point removed.

Indeed, $Ae\_1 \in \operatorname{span}(e\_1)$ and $A \in \text{SL}\_2(\mathbb{R})$ imply that

$$ A=\begin{pmatrix} \lambda & y \\\ 0 & \lambda^{-1} \end{pmatrix} \, \, \, \text{for some }\, \lambda \neq 0.$$ $A$ is conformal if and only if $\lambda=\pm 1$ and $y=0$. So, $A$ is **not** conformal if f $\lambda \neq 1,-1,0$ or $\lambda=\pm 1$ and $y \neq 0$. Thus, one connected component of $\mathcal{F}$ is homeomorphic to

$$\{ 0<\lambda \neq 1\} \times \mathbb{R} \cup \{1\} \times \mathbb{R}\setminus\{0\}.$$

(The second component corresponds to $\lambda <0$.)

---

*Here is what I know about the topology of $\mathcal{NC}$:*

Let $\mathcal D=\{ (\sigma\_1,\sigma\_2) \, | 0 \le \sigma\_1 < \sigma\_2\}$. Then the map

\begin{align\*}
\mu: SO\_2\times \mathcal D\times SO\_2\to \mathcal{NC}\\
(U,\Sigma,V)\mapsto U\Sigma V^T
\end{align\*}

is a $2$-fold smooth covering map\*. (i.e. $\mu(U,\Sigma,V)=\mu(-U,\Sigma,-V)$, and this is the only ambiguity in $U,V$ for a pre-image of a given point in $\mathcal{NC}$. 

Since $SO\_2 \cong \mathbb{S}^1$, and since [after identifying antipodal points](https://math.stackexchange.com/questions/3561814/what-is-the-surface-obtained-by-identifying-antipodal-points-of-mathbbs1-t/3563215#3563215) in $\mathbb{S}^1 \times \mathbb{S}^1$, we get the $2$-torus $\mathbb{T}^2$ again, it follows that $\mathcal{NC} \cong \mathbb{T}^2 \times D$.

\*I am not entirely sure regarding the behaviour at the boundary points where $\sigma\_1=0$, but I don't think this creates a serious problem.