TITLE: The homology of $\Omega T^n$
QUESTION [9 upvotes]: As part of a bigger plan for conquering Europe, I have to compute the integral homology of the loop space of the $n$-torus $T^n = S^1\times \cdots \times S^1$.
The plan is: compute $H_*(\Omega T^n,\mathbb{Z})$ via Serre spectral sequence (spectral sequences are part of the standard training of any supervillain) using $\Omega T^n\to pt\to T^n$ and then double-check the result using Künneth theorem.
The problem is that $\Omega T^n$ is a wicked space.
Serre spectral sequence works well when the base of a fibration is simply connected, and this is not the case. It is obviously possible to adapt the argument to a $\pi_1(B)\neq 0$ case, but (in the most powerful theorem I know) the fiber $F$ has to be connected, which is not the case ($\Omega T^n$ seems to have many${}^{\mathbb Z}$ connected components).
I tried everything I know, so please help me:

Shall I forsake my evil plan?
Is it possible to compute $H_*(\Omega T^n,\mathbb{Z})$ using Serre SS?
Shall I trust a lazy Kunneth computation which tells me that all $\Omega T^n$ have a countable number of contractible connected components?

REPLY [7 votes]: The Serre spectral sequence is an imperfect tool for calculating the homology of the fiber when the base space is not simply connected. Then you have to use (co)homology with local coefficient systems and you can't get the inductive methods started, because these types of (co)homology do not give you enough information. 
In your case, it's instructive to look at $n=1$, so your base is $S^1$. Then the homology groups of the fiber have an action of $\Bbb Z$, so we can just think of that as the generator $t$ of $\Bbb Z$ acting. All that we learn from the Serre spectral sequence is that:

the operator $t-1$ is an isomorphism on all positive degree homology groups of the fiber, and 
on the degree zero homology group, the operator $t-1$ is injective with cokernel $\Bbb Z$. 

That's simply not enough information to recover the homology of the fiber.