Long exact sequence of reduced homology

I am trying to show that the homology of suspension of a space is the homology of the space shifted by -1. I am struggling with the last part, the bottom of the long exact sequence that Mayer-Vietoris theorem gives. I found a post about someone having the same issue as me here, with someone giving a hint that if I assume it is true, then I understand the exercise is done. My problem is that I do not understand why the following is true:

We have a long exact sequence whose button is $$0\rightarrow H_1(S(X))\rightarrow H_0(X)\rightarrow Z \oplus Z\rightarrow Z\rightarrow 0$$. Then, according to the hint, the long exact sequence of the reduced homology gives $$0 \rightarrow \tilde{H}_1(S(X)) \rightarrow \tilde{H}_0(X) \rightarrow 0$$.

Why is that last statement about the reduced homology true? I have tried to a theorem that gives a long exact sequence of reduced homology given a space and subset, but I do not see the relation

As you say in the title, we deal with reduced homology $$\tilde H_n$$. We have $$\tilde H_n(Y) = H_n(Y)$$ for $$n > 0$$ and $$\tilde H_0(Y) \approx \ker (p_* : H_0(Y) \to H_0(*) )$$, where $$p : Y \to *$$ is the unique map to the one-point space $$*$$. There is a long exact Mayer-Vietoris sequence $$\ldots \to \tilde H_n(Y) \to \tilde H_n(A \cap B ) \to \tilde H_n(A) \oplus \tilde H_n(B)) \to \tilde H_n(Y) \to H_{n-1}(A \cap B ) \to \ldots$$

With $$Y = SX, A = C_+X, B = C_-X$$ we get $$\tilde H_n(A) = \tilde H_n(B) = 0$$ for all $$n$$. Thus

$$H_1(SX) = \tilde H_1(SX) = \tilde H_0(X) .$$

Edited on request:

Choose $$y \in Y$$. Since $$\tilde H_0(\{y\}) = 0$$, the long exact sequence for reduced homology ends with $$0 \to \tilde H_0(Y) \stackrel{j_*}{\rightarrow} H_0(Y,\{y\}) \to 0$$ This means that $$j_*$$ is an isomorphism. Moreover, the long exact sequence for homology ends with $$H_0(\{y\}) \stackrel{i_*}{\rightarrow} H_0(Y) \stackrel{j_*}{\rightarrow} H_0(Y,\{y\}) \to 0$$ We have $$p_* \circ i_* = (p \circ i)_* = id_* = id$$, thus $$i_*$$ is injective and we get a split short exact sequence

$$0 \to H_0(\{y\}) \stackrel{i_*}{\rightarrow} H_0(Y) \stackrel{j_*}{\rightarrow} H_0(Y,\{y\}) \to 0$$ The splitting is given by $$p_*$$. Thus $$\phi : H_0(Y) \to H_0(\{y\}) \oplus H_0(Y,\{y\}), \phi(g) = (p_*(g),j_*(g))$$ is an isomorphism. Let $$p_1 : H_0(\{y\}) \oplus H_0(Y,\{y\}) \to H_0(\{y\}), p_1(a,b) = a$$, and $$i_2 : H_0(Y,\{y\}) \to H_0(\{y\}) \oplus H_0(Y,\{y\}),i_2(b) = (0,b)$$. We have $$\ker(p_1) = \text{im}(i_2)$$. Since $$p_1 \circ \phi = p_*$$, we get $$ker(p_*) = \ker(p_1 \circ \phi) = \text{im}(\phi \circ i_2) \approx H_0(Y,\{y\}) \approx \tilde H_0(Y)$$.

• Would you mind clarifying why we have the isomorphism $\tilde H_0(Y) \approx \ker (p_* : H_0(Y) \to H_0(*) )$ ? Mar 11, 2020 at 3:14
• Have a look at my edit. Mar 11, 2020 at 12:35
• Thank you! I appreciate your help Mar 11, 2020 at 15:20
• By the way, for your question we do not need the fact that $\tilde H_0(Y) \approx \ker (p_* : H_0(Y) \to H_0(*) )$. We can directly see that $\tilde H_0(*) = 0$ which implies that also $\tilde H_0(C_\pm X) = 0$. Mar 12, 2020 at 12:55