Let
$$ X^{\square\setminus\Delta}\ :=\ \{(x\ y)\in X^2 : x\ne y\} $$
be the deleted square of X. When $\ X\ $ is a topological space, then the deleted square (with the subspace topology induced by the square) is a topological invariant which is not a homotopy invariant. Then several other derived operations are likewise topologically but not homotopically invariant even when they are a composition of the deleted square and of a homotopically invariant operation--for instance, $\ \pi_1(X^{\square\setminus\Delta})\ $ is topologically but not homotopically invariant. This worked well for manifolds (in particular in Hirsch's hands).
EXAMPLE Let $\ I\ $ be a closed interval, and $\ T\ $ be any finite tree with more than two endpoints; thus, $\ I\ $ and $\ T\ $ are homotopically equivalent (to the 1-point space). However $\ T^{\square\setminus\Delta}\ $ is connected, while $\ I^{\square\setminus\Delta}\ $ is disconnected.
We see that the topologically invariant operation of the deleted square, as well as the fundamental group of the deleted square, can distinguish between homotopically invariant spaces (even in the simple cases).