Homology groups equal when $\cdots \rightarrow 0 \rightarrow H_n(\mathbb{S}^m) \rightarrow H_{n-1}(\mathbb{S}^m) \rightarrow 0 \rightarrow \cdots$

Question

I'm reading a set of notes but I don't understand the following concept. We have a long exact sequence

$\cdots \rightarrow 0 \rightarrow H_n(\mathbb{S}^m) \rightarrow H_{n-1}(\mathbb{S}^m) \rightarrow 0 \rightarrow \cdots$

Apparently this implies that $H_n(\mathbb{S}^m)$ is isomorphic to $H_{n-1}(\mathbb{S}^m)$. Why is this true? I tried to prove it using exactness and rank nullity but all I could show is that rank$(H_n(\mathbb{S}^m)) = $ rank$(H_{n-1}(\mathbb{S}^m))$

Answer 1

It might be clearer if you actually name the maps. You have a short exact sequence within the long exact sequence $$ 0\xrightarrow{f}H_n(S^m)\xrightarrow{g} H_{n-1}(S^m)\xrightarrow{h} 0 $$ where $f$ and $h$ are zero maps. Then by exactness, $0=\operatorname{im}(f)=\ker g$, so $g$ is injective. Also, $\operatorname{im}(g)=\ker h=H_{n-1}(S^m)$ since $h\equiv 0$. Thus $g$ is surjective, hence an isomorphism.

There is no need to look at the ranks of the groups.

Answer 2

If the coefficients in the sequence are in some field, then you have a linear map between two vector spaces of the same rank. By exactness, the kernel of the map is trivial, so the map is onto. Now from rank-nullity theorem, the image is n-dimensional subspace of n-dimensional space. Alternatively, if not working over a field. then, by exactness ,the kernel of the $0$ map is $H_{n-1}$, i.e., the whole space, which coincides with the image o the map, so the map is onto 1-1 and linear, i.e., an isomorphism.