Embeddings into products and symmetric products -- an algebraic approach (Q2901926)

Every separable metrizable space \(X\) of dimension \(\leq n\) embeds into \(\mathbb R^{2n+1}\), and \(\mathbb R^{2n+1}\) is the topological product of one dimensional factors which are all \(\mathbb R\). The exponent \(2n+1\) can not be lowered in general. Furthermore, \(\mathbb R^{2n+1}\) contains an \(n\)-dimensional subset \(S\subset \mathbb R^{2n+1}\) of dimension \(n\) (universal space) and each \(X\) of dimension \(\leq n\) embeds into \(S\). If in the topological product we admit different 1-dimensional factors, then \textit{J. Nagata} [Fundam. Math. 45, 143--181 (1958; Zbl 0082.16303)] has shown that \(X\) embeds into a topological product \(Y_1\times\dots \times Y_{n+1}\) of 1-dimensional spaces. \textit{K. Borsuk} [Bull. Acad. Pol. Sci., Sér. Sci. Math. Astron. Phys. 23, 971--973 (1975; Zbl 0312.57001)] added that \(n+1\) can not be lowered in general by showing that \(S^n, n\geq 2,\) can not be embedded in a product of \(n\) curves (one dimensional continua). In the present paper it is proved that \(S^n\), \(n\geq 2,\) can not be embedded in the \(n\)th symmetric product of any curve.NEWLINENEWLINEThe paper under review studies algebraic properties of certain compacta lying either in products of \(n\) 1-dimensional curves or in the \(n\)th symmetric product of a curve. In the next step compacta are considered lying in products \(X\subset \prod_1^k Y_i\), such that \(\dim X =\sum_1^k \dim Y_i =n\geq 1\), what finally comes to \(X\subset Y\) \(\dim X = \dim Y =n\geq 1\). A series of new results and generalizations of some known ones is obtained.NEWLINENEWLINELet us illustrate the results of this paper by listing three simplified versions of the obtained results:NEWLINENEWLINE1. Let \(X\) be a compact subset of the product of \(n\) curves, \(n>1,\) and \(G\) an Abelian group such that \(H^n(X;G)\not =0\). Then \(H^1(X;G) \not= 0.\)NEWLINENEWLINE2. Let \(X\) be a closed subset of the product \(Y_1\times \dots \times Y_n\) of \(n\) curves with \(H^n(X)\not= 0\). Then there is an algebraically essential map from \(X\) into the \(n\)-torus \(T^n\). Consequently, there exist elements \(a_1, \dots , a_n \in H^1(X)\) whose cup product \(a_1\smile \dots \smile a_n\) belonging to \(H^n(X)\) is non-zero. Such elements are linearly independent, hence rank \(H^1(X)\geq n.\) Moreover, cat \(X>n\), where cat \(X\) denotes the Lusternik-Schnirelmann category of \(X\).NEWLINENEWLINE3. If \(Y\) is an \(n\)-dimensional compactum, \(n\geq 1\), \(G\) an Abelian group, \(H\) a proper subgroup of \(H^n(Y;G)\), then there exists a map \(\varphi: Y\to S^n\) such that the image im \(H^n(\varphi;G)\) is not a subset of \(H\).