Example

Let G be the subgroup of $\mathbf{G}\mathbf{L}(n,\mathbf{R})$ formed by the square matrices of order $n$ whose determinant is strictly positive. The group G is connected and locally path-connected.

According to A, III, p. 104, prop. 17, the group $\mathbf{S}\mathbf{L}(n,\mathbf{R})$ is generated by the elements $B_{ij}(\lambda )$ (for $1\leqslant i, j\leqslant n$ such that $i=\not j$ and $\lambda \in \mathbf{R}$). The mappings $\lambda \mapsto B_{ij}(\lambda )$ from $\mathbf{R}$ into $\mathbf{G}\mathbf{L}(n,\mathbf{R})$ are continuous, their images are connected subsets of $\mathbf{S}\mathbf{L}(n,\mathbf{R})$; since they all contain the identity matrix $I_n$, their union is connected (TG, I, p. 81, prop. 2). Consequently, the group $\mathbf{S}\mathbf{L}(n,\mathbf{R})$ is connected (TG, III, p. 8, prop. 7). Let A be the subgroup of G formed by the matrices of the form diag(1$, . . . ,1, \lambda$ ), with $\lambda \in \mathbf{R}_+^*$; it is connected and one has $\mathbf{G}\mathbf{L}(n,\mathbf{R}) = A\cdot \mathbf{S}\mathbf{L}(n,\mathbf{R})$. It follows that the group G is connected. Since it is the inverse image of $\mathbf{R}_+^*$ by the determinant mapping of $\mathbf{M}_n(\mathbf{R})$ into $\mathbf{R}$, it is an open subset of $\mathbf{M}_n(\mathbf{R})$; it is therefore locally arcwise connected (corollary above), as well as arcwise connected (III, p. 261, corollary 1).