No, such a space ( X ) cannot exist. If ( X \times Y ) is homotopy equivalent to a finite-dimensional CW complex, then ( X ) itself must be homotopy equivalent to a finite-dimensional CW complex. Here's the detailed reasoning:

Key Observations:

1. Finite-Dimensional CW Complexes: A space ( Z ) homotopy equivalent to a finite-dimensional CW complex has finite cohomological dimensio...

No. If X × Y has the homotopy type of a finite-dimensional CW complex, then X and Y each do as well.

Here is a clean argument.

• Let Z ≃ X × Y be homotopy equivalent to an m-dimensional CW complex. Then for every local coefficient system L on Z we have H^i(Z; L) = 0 for i > m, since cohomology with local coefficients is computed by a length-m cochain complex.

• Fix any local coefficient system ...

Here is the one‐line “no‐go’’ argument which shows that you cannot manufacture an exotic counter–example by multiplying by some clever space (Y). In fact:

(1) Fix a point (y_0\in Y). Then [ i\colon X;\longrightarrow;X\times Y,\quad x;\mapsto;(x,y_0) ] is a (strict) section of the projection [ p\colon X\times Y;\longrightarrow;X,\q...

This is a fascinating question about when products of spaces can have better finiteness properties than their factors. Let me provide some insights and a construction that addresses your question.

The Answer is Yes

There exist spaces that are not homotopy equivalent to finite-dimensional CW complexes, but for which there exists such that is homotopy equivalent to a finite...

No.
If a product X × Y has the homotopy type of a finite–dimensional CW–complex, then each factor already has the homotopy type of a finite-dimensional CW–complex (in fact of dimension not larger than that of the product). Hence no counter–example can exist.

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Outline of the argument --------------------------------...

<think> I need to figure out if there's a space X that isn't homotopy equivalent to a finite-dimensional CW complex, but when multiplied by some other space Y, the product X × Y is homotopy equivalent to a finite-dimensional CW complex. The question mentions that if X has infinitely many nontrivial homology groups, then no such Y exists because of the Künneth formula. Also, for Eilenberg-MacLane s...

The phenomenon you hope for cannot occur:

If the product X × Y has the homotopy type of a finite-dimensional CW–complex, then each factor X and Y already has the homotopy type of a finite-dimensional CW–complex (in fact of dimension ≤ dim(X × Y)). Hence no counter–example exists.

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1. From “finite dimensional product’’ to “fi...

This is an excellent question that touches on a fundamental property of topological spaces related to their dimension. The short answer is no, such a space does not exist, at least under the very weak assumption that and have the homotopy type of CW complexes.

The property of being homotopy equivalent to a finite-dimensional CW complex is "hereditary" for factors in a product. Let...

If a Product Space is Homotopy Equivalent to a Finite-Dimensional CW Complex, are its Factors Also?

This is an interesting topological question about when product spaces and their factors have finite-dimensional CW structure. Let me provide a thorough analysis.

The Short Answer

No, there is no space that is not homotopy equivalent to a finite-dimensional CW complex for which there exis...