Let $H_i$ be the $i$-th homology group of a topological space $X$ ($X$ to be read from context). The Wikipedia page on homology says the circle $S_1$ has $H_1$ equal to $\mathbb{Z}$, so that it "has" a "one-dimensional hole". Furthermore, $S_2$ has $H_2$ equal to $\mathbb{Z}$ because it has a two-dimensional hole, and $S_n$ has $H_n$ equal to $\mathbb{Z}$ because it has an n-dimensional hole.
I'm trying to come up with a naive interpretation of what it means to "have" an $n$-dimensional hole. $S_1$, embedded on the plane, encloses (ie. is the boundary of) a $2$-ball (aka. disk), $S_2$ encloses an $3$-ball, etc., so my naive interpretation would be, at any rate, that $S_n$ "has" an $(n+1)$-dimensional hole (which is false).
I'm sure this question can be answered with a little understanding of elementary homological machinery, but, for now: what is an intuitive explanation of an $n$-dimensional manifold "having" a $k$-dimensional hole?
(An $i$-dimensional "hole" --aka. a homology class-- seems to (somehow) be an element of $H_i:=\text{Ker }d_i /\text{Img }d_{i+1}$, which I think measures how much the underlying chain complex fails to be an exact sequence, since chain complexes satisfy $\text{Img }d_{i+1} \subseteq \text{Ker }d_i$ whereas an exact sequence would have equality.)
(This question doesn't provide much information.)