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I am confused by the terminology concerning $n$-dimensional holes in algebraic topology. A circle is said to have a one-dimensional hole, and a sphere a two-dimensional hole for example. However I cannot see why the circle should be described to have a one-dimensional hole $-$ surely if drawn in two dimensions the 'gap' left in the middle of the circle is two-dimensional? If we think of the circle as a one-dimensional space only, then there is nowhere to have a 'hole'?

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    What is your definition of $n$-dimensional hole? – 2017-02-06
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    It's not about the dimension of the "filling", because that's not part of the space itself, and therefore in some sense dependent on exactly how you're picturing the space. Instead, the dimension of a hole is, intuitively, the dimension of the natural entity you would use, within the space itself, to enclose / encircle the hole. That way, a single point in $n$-dimensional Euclidean space is an $(n-1)$-dimensional hole. – 2017-02-06
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    @Watson this is a good point. I see the term used but don't ever think I saw it formally defined. I am new to the field. – 2017-02-06
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    Do you know homology? – 2017-02-06
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    @Watson I am just starting out. I hope to be self-studying it as part of preliminary PhD studies. – 2017-02-06
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    The fact that $H_n(S^n) \cong \Bbb Z$ somehow indicates that there is an $n$-dimensional hole in $S^n$. – 2017-02-06
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    If people stopped using this "hole" idea people would learn faster and better... The ideas of bounding and of cycles being homologous, which **precisely** describes what homology is, is the correct intuitive point of view: I cannot tell why everyone seems to avoid it. – 2017-02-06
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    Why on earth would this be downvoted? It is something obvious I am missing? – 2017-02-07
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    To expand my previous comment: $H_1(S^1)$ is the quotient of the 1-cycles by the 1-boundaries. The fact $H_1(S^1) \cong \Bbb Z$ means that there is a 1-cycle which is not a boundary. This is confirmed by intuition: the circle $S^1$ is indeed a cycle, but it is _not_ the boundary of something else. Here the fact that $S^1$ is not a boundary means that there is some "hole" which prevents the cycle $S^1$ to be a boundary. On the other hand, if you consider the disk $D^2 \subset \Bbb R^2$, then $S^1$ _is_ a boundary — here we have $H_1(D^2)=0$. – 2017-02-12

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