I have been working through the following exercise:
Consider $X = GL (n, \mathbb{C})$ with respect to the Euclidean topology in $\mathbb{C}^{n^2}$.
Find a subspace $Y \subset X$ such that $Y$ is homemorphic to $S^1$ and $Y$ is a retract of $X$.
Show that the fundamental group $\pi_1 (X, x)$ is infinite for all $x \in Y$.
Let's consider the set $Y$ which consists of all matrices \begin{equation*} \begin{pmatrix} z & 0 & \ldots & 0 \\ 0 & 1 & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \ldots & 1 \end{pmatrix}, \end{equation*} with $z \in \mathbb{C}$ such that $|z| = 1$.
Then $Y$ is a subspace of $X$ and it is homeomorphic to $S^1$.
Let $r : X \rightarrow Y$ be the map defined by \begin{equation*} r(A) = \begin{pmatrix} \frac{\det A}{|\det A|} & 0 & \ldots & 0 \\ 0 & 1 & \ldots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \ldots & 1 \end{pmatrix} \end{equation*} for all matrices $A \in X$. This map is continuous because the determinant is a continuous map and every matrix in $X$ has non-zero determinant. Besides, $r$ is a retraction of $X$ to $Y$, because $r|_Y$ is the identity map on $Y$.
Let's consider the inclusion $i : Y \rightarrow X$ and a point $x \in Y$. Since $r$ is a retraction, the group homomorphism \begin{equation*} i_* : \pi_1 (Y, x) \rightarrow \pi_1(X, x) \end{equation*} is injective. Then $\pi_1(X, x)$ is infinite, because $\pi_1 (Y, x)$ is an infinite (cyclic) group.
Is this solution acceptable?