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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

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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 $ker(p_*) = \ker(p_1 \circ \phi) = \text{im}(\phi \circ i_2) \approx H_0(Y,\{y\}) \approx \tilde H_0(Y)$.

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  • $\begingroup$ Would you mind clarifying why we have the isomorphism $\tilde H_0(Y) \approx \ker (p_* : H_0(Y) \to H_0(*) )$ ? $\endgroup$ Mar 11, 2020 at 3:14
  • $\begingroup$ Have a look at my edit. $\endgroup$
    – Paul Frost
    Mar 11, 2020 at 12:35
  • $\begingroup$ Thank you! I appreciate your help $\endgroup$ Mar 11, 2020 at 15:20
  • $\begingroup$ 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 $\tilde H_0(C_\pm X) = 0$. $\endgroup$
    – Paul Frost
    Mar 12, 2020 at 12:55

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