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True. The number of holes is typically associated with the Betti numbers of a space, which are homotopy invariants. This means that if two spaces are homotopy equivalent, they must have the same number of holes.

True. Homology groups are homotopy invariants, which means that if two spaces are homotopy equivalent, their homology groups should be isomorphic. This property makes homology groups particularly useful in algebraic topology for classifying spaces up to homotopy equivalence.

False. The fundamental group is a homotopy invariant, which means that if two spaces are homotopy equivalent, they must have isomorphic fundamental groups. Therefore, if two spaces have different fundamental groups, they cannot be homotopy equivalent.

False. While the entire space is contractible, a retract might not be. For example, consider a disk with a smaller disk removed from the interior (like an annulus). The disk is contractible, but the annulus is not, although it is a retract of the larger disk.

True. Given two pairs of homotopy equivalent spaces $X \sim X'$ and $Y \sim Y'$, we have that their products $X \times Y$ and $X' \times Y'$ are also homotopy equivalent. This follows because the maps demonstrating the homotopies between $X$ and $X'$ and $Y$ and $Y'$ can be combined to form a homotopy between $X \times Y$ and $X' \times Y'$.

True. Homeomorphic spaces are topologically the same, which means there exist continuous functions that map one space onto the other with continuous inverses. Since homotopy equivalence is a more relaxed condition, requiring only that the spaces can be deformed into each other continuously (without necessarily preserving points exactly), homeomorphism implies homotopy equivalence.

True. The cone over a space $X$ is created by adjoining an extra dimension and collapsing one boundary to a point. This new space can always be contracted to the tip of the cone, which is the image of the collapsed boundary, demonstrating that the cone is contractible.

False. A circle is not contractible, as it cannot be continuously deformed to a point without cutting the space. This is reflected by the fact that the fundamental group of the circle is non-trivial (it is isomorphic to $\mathbb{Z}$), while the fundamental group of a point is trivial. Therefore, a circle is not homotopy equivalent to a point.

True. Convex subsets of $\mathbb{R}^n$ are contractible since any point can be connected to any other point via a straight line segment (which lies entirely in the convex set). Thus, all convex subsets can be contracted to a point and hence are homotopy equivalent to each other.