Higher homology groups relative a lower dimensional subspace One often works with reduced homology, which (in the case of say, smplicial homology) is defined as the homology relative a point. Now at every grade except zero, the reduced homology objects are isomorphic to the original ones.
Intuitively, I think this is because a point is $0$-dimensional, so it can only interfere with $0$-cycles. Now suppose I'm looking at the relative homology objects $H_n(X,A)$. Suppose $A$ is a $k$-dimensional manifold in the space $X$. The same intuition temps me to assert that for all $n>k$ $$H_n(X,A)\cong H_n(X)$$
Is this true? How can I prove it? Can this be somehow generalized to topological dimension?
 A: Relative homology is more or less defined so that it fits into a long exact sequence
$$\dots \xrightarrow{\partial} H_n(A) \to H_n(X) \to H_n(X, A) \xrightarrow{\partial} H_{n-1}(A) \to \dots$$
which straightforwardly answers your question: the natural map $H_n(X) \to H_n(X, A)$ is an isomorphism iff it has trivial kernel and full image. By exactness, its kernel is the image of $H_n(A) \to H_n(X)$; a sufficient but not necessary condition for this image to be trivial is that $H_n(A)$ vanishes. By exactness, its image is the kernel of $H_n(X, A) \to H_{n-1}(A)$; again a sufficient but not necessary condition for this kernel to be full is that $H_{n-1}(A)$ vanishes. 
So a sufficient but not necessary condition is that $A$ has homological dimension less than $n - 1$. Your intuition should be corrected as follows: the long exact sequence shows that $H_n(A)$ can interfere with both $n$-cycles and $n+1$-cycles, and intuitively this is because on the one hand it can remove some $n$-cycles and on the other hand it can add some $n+1$-cycles, roughly because it can cause some $n+1$-chains to become cycles that weren't before. 
