In singular homology one of the first calculations you can make is $H_0(X)=H_0(pt)$ for path-connected $X$. This seems to be a property which does not follow from the axioms for a generalized homology theory. This raises the following question:
Assume $H_* : Top^2 \to Ab^\mathbb{N}$ is a homology theory. Thus we impose homotopy invariance, excision, the long exact sequence, the dimension axiom and if you wish also the disjoint union axiom. Is then $H_0(X)=H_0(pt)$ for every path-connected space $X$?
I believe that there is a counterexample. Of course this can't be homotopy equivalent to a CW-complex. And probably this is the reason why this question is not reasonable at all. It's just my curiosity.
Best Answer
Here is a candidate space for Cech homology. (which doesn't satisfy the disjoint union axiom, and so strictly speaking it's not a homology theory)
For any natural number n, let $X_n$ be the circular arc of radius 1 + 1/n centered at $( 1+1/n,0)$ and extending between angles $-\pi \leq \theta \leq (\pi - 1/n)$. Let $X$ be the union of these circles with the subspace topology, which is path connected.
Covering this space with sufficiently small disks (each $X_n$ needing to be covered by progressively smaller disks) gives a Cech nerve homotopy equivalent to a similar space where all but finitely many of the circular arcs have been closed up to circles. The zero'th homology of this cover is $\mathbb{Z}$ and the first homology is an infinite direct sum $\oplus_{n \geq N} \mathbb{Z}$. All other homology groups are zero.
As you decrease the size of the cover, you get a cofinal sequence of open covers inducing a decreasing sequence of abelian groups as N grows. There is a resulting exact sequence $$ 0 \to lim^1(\oplus_{n \geq N} \mathbb{Z}) \to \check{H}_0(X) \to lim^0(\mathbb{Z}) \to 0 $$ and the left-hand side is $(\prod \mathbb{Z}) / (\oplus \mathbb{Z}) \neq 0$.