TITLE: Lusternik-Schnirelmann category: nullhomotopic inclusion vs. contractible
QUESTION [8 upvotes]: The Lusternik-Schnirelmann category of a topological space $X$ is the smallest integer $k$ (if it exists) such that there is an open cover $\{U_0, \dots, U_k\}$ of $X$ such that each inclusion map $U_i \hookrightarrow X$ is nullhomotopic; we denote this by $LS(X) = k$. If no such integer exists, we write $LS(X) = \infty$.
It should be noted that some references (e.g. Wikipedia) use a different normalisation which differs from the above by one.
The Lusternik-Schnirelmann category is a homotopy invariant which enjoys has several nice properties, making it an interesting invariant to study. I wonder, however, whether there is any difference between using an open cover by sets where the inclusions are nullhomotopic (sets which are sometimes called 'contractible in $X$') and using an open cover by contractible sets.
Define the alternative Lusternik-Schnirelmann category of a topological space $X$ to be the smallest integer $k$ (if it exists) such that there is an open cover $\{U_0, \dots, U_k\}$ of $X$ such that each $U_i$ is contractible; we denote this by $LS'(X) = k$. If no such integer exists, we write $LS'(X) = \infty$.

Question 1: Is $LS'(X)$ a homotopy invariant? 

If $LS'(X)$ is a homotopy invariant, we can compare the invariants $LS(X)$ and $LS'(X)$. If $U \subseteq X$ is contractible, the inclusion $U \hookrightarrow X$ is nullhomotopic, so it follows that $LS(X) \leq LS'(X)$. 

Question 2: Is $LS(X) = LS'(X)$?

REPLY [3 votes]: $LS'$ is not a homotopy invariant. It is straightforward to verify that $LS(X \vee Y) = \text{max}(LS(X),LS(Y))$ for reasonable spaces (we need to demand that in our minimal open cover, an open set containing the basepoint has the basepoint component deformation retract onto the basepoint), and $LS'(X \vee Y) = LS'(X) + LS'(Y)$. (I can provide proofs if desired.) In particular, $LS'(\Sigma T^2) = 1$, since we can cover it by the two cones, but this is homotopy equivalent to $S^3 \vee S^2 \vee S^2$, and $LS'(S^3 \vee S^2 \vee S^2) = 3$. The important point is that for $LS$, the open sets in your cover can be disconnected, while this is not the case for $LS'$. Your definition stands more of a chance of survival if you define $LS'$ in terms of covers by open sets with contractible components  (but this is no longer so exciting an alternate definition).