$\textbf{Definition:}$ Let $X$ be a Banach space and $A \subset X \backslash \{ 0 \}$ a closed set and symmetric with respect to the $0 \in X$. The Krasnoselkii's genus of $A$ is defined as follows
$$\sigma(A) := \min \{ m \in \mathbb{N} ; \ \text{exists an odd and continuous function} \ \varphi: A \rightarrow \mathbb{R}^m \backslash \ \{ 0 \} \}.$$
$\textbf{Clark's theorem:}$ Let $X$ be a Banach space and let $J \in C^1(X,\mathbb{R})$ be a functional satisfying Palais-Smale condition. Furthermore, suppose that
$(i)$ $J$ is limited from below and even;
$(ii)$ Exists a compact $K$, which is possible compute the Krasnoselkii's genus, such that the $\sigma(K) = j$ and $\sup_\limits{x \in K} J(x) < J(0)$,
then $J$ has, at least, $j$ even critical points distincts and their corresponding critical values are less than $J(0)$.
I would like to understand a part of the application of this theorem for this problem:
$$\begin{align} \begin{cases} - \Delta u &= |u|^{q-2}u \ \text{in} \ \Omega,\\ u &= 0 \ \text{on} \ \partial \Omega, \end{cases} \end{align}$$
where $1 < q < 2$ and $\Omega \subset \mathbb{R}^N$ is a bounded domain.
The functional associated to the problem is $J(u) := \frac{1}{2} \int_{\Omega} |\nabla u|^2 dx - \frac{1}{q} \int_{\Omega} |u|^q dx, u \in H_0^1(\Omega)$.
Define $X_j := \text{span} \{ \varphi_1, \cdots, \varphi_j \}$, where $\varphi_i$ is the eigenfunction of $-\Delta$ associated to the eigenvalue $\lambda_i$ for each $i = 1, \cdots, j$. As $X_j$ is a normed space of finite dimension, there is $C_j > 0$ such that
$$- \frac{C_j}{q} \left( \int_{\Omega} |\nabla u|^2 dx \right)^{\frac{q}{2}} > - \frac{1}{q} \int_{\Omega} |u|^q dx, \forall u \in X_j$$
Thus, if $||u|| = \rho$ with $\rho$ sufficiently close to $0$, the inequality above implies that
$$J(u) \leq \frac{1}{2} \int_{\Omega} |\nabla u|^2 dx - \left( \int_{\Omega} |\nabla u|^2 dx \right)^{\frac{q}{2}} < 0 = J(0), \forall \partial B_{\rho}(0) \cap X_j.$$
Define $K := B_{\rho}(0) \cap X_j$.
I know that $X_j$ is isomorphic to $\mathbb{R}^j$, but how I can construct an odd homeomorphism between $K$ and $S^{j-1}$ in order to conclude that $\sigma(K) = j$?
Thanks in advance!