# Why is this isomorphic to the reduced homology group?

I am looking at the Mayer-Vietoris sequence for the suspension $\Sigma X$ of a space $X$, defined as $X\times [-1,1]$ with the usual identifications, with subsets $A, B$ defined to be the following $$A = X \times [-\tfrac{1}{2},1]/\sim,$$ $$B = X \times [-1,\tfrac{1}{2}]/\sim,$$ where $\sim$ is just the equivalence relation inherited from $\Sigma X$. Only the last part of Mayer-Vietoris is important to my question: $$0 \to H_1(\Sigma X) \to H_0(X)\: \xrightarrow{f_*} \: H_0(A) \oplus H_0(B) \to H_0(\Sigma X) \to 0.$$ Since $A$ and $B$ are contractible (they are homeomorphic to the cone on $X$, which is contractible), $H_0(A) \oplus H_0(B) \cong \mathbb{Z} \oplus \mathbb{Z}$.

Question: I am then given that $H_1(\Sigma X) \cong \ker f_* \cong \tilde{H}_0(X)$. The second isomorphism is the problem; I thought that the reduced $0$th homology was defined to be $\ker \big(g_*:H_0(X) \to H_0(pt)\big)$, where the codomain is the homology of a point. Isn't this different to what we have here?

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Identify $H(A)$ and $H(B)$ with $H(pt)$. Then $f_*=(g_*,-g_*)$, so $\ker f_*=\ker g_*$.
Just to correct my intuition, is the map $g_*$ that is used to define reduced homology the zero map for all $n$th homology groups with $n>0$, and the identity map for $n=0$? –  Sputnik May 29 '11 at 18:34
@Fahad Yes, it's always zero for $n>0$ (since $H_n(pt)=0$) and for $n=0$ it's iso whenever $X$ is connected. –  Grigory M May 29 '11 at 18:37
Okay. Then supposing $X$ is not connected, what does $g_*$ actually do as a function? Is $g_*([\sigma]) = [\sigma]$? –  Sputnik May 29 '11 at 18:53
@Fahad I don't quite understand last question. But, anyway, hint: $g$ can be factored as $X\to\pi_0(X)\to pt$. –  Grigory M May 31 '11 at 4:44