Let $X\subset\mathbb{R}^{\infty}$ be a paracompact connected Hilbert manifold. Moreover, let $f:X\to\mathbb{R}$ be a $C^2$-function.
Then is the simplicial complex $\Delta:=\{p\in X:df(p)=0\}$, with vertices the non-degenerate critical points of $f$ (of which, there may be countably or uncountably infinitely many), homotopy equivalent to $X$?
Note, a $C^0$-perturbation of $f$ may produce infinitely many critical points. Then suppose we form a good cover $\{U_{\alpha_k}\}_{\alpha_k\in A}$ such that each $U_{\alpha_k}$ is a neighborhood of $p_k$. Then, in general, this should be a cover of $X$ and the map from the geometric realization of the simplicial space (i.e. the topological nerve of $\{U_{\alpha_k}\}$) to $X$ is a homotopy equivalence. Likewise, the geometric realization $R(\text{Top-Nerve}({U_{\alpha_k}}))$ is homotopy equivalent to the realization of the simplicial set for which the $\infty$-simplices are $\infty$-tuples $(\alpha_0,\dots)$ whereby the intersection $\bigcap_{k\in\mathbb{Z}}U_{\alpha_k}$ is nonempty. Is this argument correct?
This seems justified if we consider the nerve of a good cover $\{U_{\alpha_k}\}_{\alpha_k\in A}$ around each critical point of which there can be infinitely many (i.e. $|A|=\infty$). However, I think that the torus $S^1\times S^1$ could be a possible counterexample.
Any help would be much appreciated. Thanks in advance!