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i have completed a basic course in algebraic topology and am currently pursuing a course in homology theory.Since the day i have started a course in homology theory and got involved in finding homology groups etc,i have always faced two problems:

(1)How do we arrive at the formulae of nth homology group (i.e,$H_n(K)=\frac{Z_n(K)}{B_n(K)}$,where, K is a simplicial complex)from it's geometrical intuition which says that nth homology denotes the presence of n dimensional hole in the space.how can we connect the geometrical intuition behind the homology groups with the formulae for the nth homology group.how does the concept of cycles modulo boundary give us information regarding the presence of hole in the space??

(2)i really get confused while trying to differentiate between the topological aspects of the surface shared by homology groups and fundamental groups.As far as i percieve from my basic knowledge of fundamental groups,the study of equivalence classes of loops or closed paths in any topological space X might be a way of determining the 'holes' in the space.So,intuitively both of them refer to the same aspect of counting holes in the spaces.So,i am really unable to find any sort of INTUITIVE differences regarding topological property of space shared by them.

i am really sorry if the above question does not meets the standard of stack exchange.

if possible,try to give detailed explanations...

thanks in advance....

Abhishek Shrivastava
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  • For 1, you might consider the simple examples of $\mathhb{R}^3$ minus a line versus $\mathbb{R}^3$ minus a point. The former has nontrivial first homology (it's not simply connected) but trivial second homology, while the latter has trivial first homology (it is simply connected) but nontrivial second homology. – symplectomorphic Mar 09 '16 at 07:58
  • The first homology is the "abelianization" of the fundamental group, meaning that while both "count holes", the fundamental group cares about the order in which a loop might travel around different "holes". Chapter 2 of Hatcher was what made homology intuition click for me; look there. – Mose Wintner Mar 09 '16 at 09:20

1 Answers1

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To see why homology is supposed to "count holes", take the case where $n=1$. Then a singular $1$-chain is just a path in $X$ (since the $1$-simplex is basically $[0;1]$). So a cycle is a sum of path so that the endpoints "telescope". For instance, if you put paths one after the other to make up a triangle, this will be a $1$-cycle. Now a $2$-boundary is the boundary (yes...) of a full triangle in $X$ (actually a sum of such things). So if you take $\mathbb{R}^2$ minus the origin, and take a $1$-cycle as described above to loop around the origin, you won't be able to find a full triangle that has that as a boundary, because the full triangle would have to include the missing point. On the other hand, if you don't have this hole, then no problem, you just take the obvious full triangle with this boundary.

As for your other question, it's true that homotopy groups and homology groups are deeply linked. For instance (let's assume $X$ is arcwise connected to avoid mentioning basepoints), $H_1(X)=\pi_1(X)^{ab}$, and more generally for $n\geqslant 2$ if $\pi_i(X)$ is trivial for $i<n$ then $H_i(X)$ is also trivial for $i<n$, and $H_n(X)=\pi_n(X)$.

The $\pi_n(X)$ are a "perfect" homotopical invariant, in the sense that for nice spaces, if $f:X\to Y$ induces isomorphisms $\pi_n(X)\to \pi_n(Y)$ for all $n$, then $X$ and $Y$ are homotopically equivalent. But they are close to impossible to compute in general.

Whereas homology is much weaker, but on the other hand is very computable. So it is a sort of poor man's homotopy group : it's less powerful, but you can actually get it. This can already be seen in $H_1(X)=\pi_1(X)^{ab}$ : when you compute $H_1$ instead of $\pi_1$, you are in a sense settling for less information.

Captain Lama
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