TITLE: Intuition of non-free homology groups?
QUESTION [16 upvotes]: According to Intuition of the meaning of homology groups and Wikipedia, homology groups can be thought of as counting the number of holes in a given dimension. For instance, a wedge of two circles has $H_1 = \mathbb{Z}^2$ because there are two holes enclosed by the space, and $S^2$ has $H_2 = \mathbb{Z}$ because it encloses a volume, and it has $H_1=0$ because any loop is homotopic to a point.
So how can I have a better understanding of spaces like the real projective space $\mathbb{R}P^n$, or the Klein bottle, whose homology groups aren't free? I understand that the Klein bottle has $H_2 = 0$ because it encloses no volume, and $H_1 \geq \mathbb{Z}$ because loops around it aren't homotopic to a point, but how does it make intuitive sense that $H_1 = \mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$?

REPLY [8 votes]: I'm not sure how to describe this precisely, but i had this question myself and this is a qualitative answer that I found intuitive:
In these spaces you can place a loop which is not trivial, but if you trace the loop twice, the result can be continuously deformed within the space to become trivial. That is the loop is it's own inverse.  
In the $\mathbb{RP}^2$ case, consider $\mathbb{RP}^2$ as the quotient space of the square in the standard way, draw a line from a corner to the opposite corner. You can convince yourself that this is a non trivial loop, but by continuously rotating the line about $180^\circ$, the opposite ends are always identified under the quotient map so we are continuously moving the loop, but we reverse the orientation, so the loop is it's own inverse. 
We can do a similar thing with the Klein bottle. Consider the Klein bottle as the cylinder, with top and bottom circles identified with opposite orientation. The loop around the top circle is then the same as the loop around the bottom circle with opposite orientation. Mapping it continuously to itself by moving it down the cylinder shows then that it is in some sense it's own inverse.