Define $\Sigma X$ to be the quotient space of $[-1,1]\times X$ obtained by identifying $\{-1\}\times X$ and $\{1\}\times X$ to two points respectively. For any homology theory (satisfying Eilenberg-Steenrod axioms), I am able to find an isomorphism $\tilde H_i(X) \rightarrow \tilde H_{i+1}(\Sigma X)$ as follows:

Denote $C_+ X=[0,1]\times X /\tilde{}$ and $C_- X=[-1,0]\times X /\tilde{}$, then we know both of them are contractible and $\Sigma X = C_+X \cup C_- X$. First we consider the reduced homology sequence for the pair $(\Sigma X, C_+ X)$: $$ 0=\tilde H_i(C_+ X) \to \tilde H_i(\Sigma X) \to H_i(\Sigma X,C_+ X) \to \tilde H_{i-1}(C_+ X)=0 $$ Hence we know $$\tilde H_i(\Sigma X) \to^{\cong} H_i(\Sigma X,C_+ X)$$ is an isomorphism. Second, by considering the reduced homology sequence for the pair $(C_- X, X)$(where X is identified with the quotient image of $\{0\}\times X$), we can similarly get $$ H_i(C_-X,X)\to^{\cong} \tilde H_{i-1}(X) $$ Finally, using excision axiom and homotopy axiom we can show that $$ H_i(C_-X,X) \cong H_i(\Sigma X,C_+ X) $$

Nevertheless, I have no idea how to show this isomorphism is also "natural". Here "natural" means that, if we denote this isomorphism by $\Phi: \tilde H_i(X) \to \tilde H_{i+1}(\Sigma X)$, then, for any map $f:X\to Y$ and its suspension $\Sigma f: \Sigma X \to \Sigma Y$, $\Phi f_* = (\Sigma f)_* \Phi$.

  • $\begingroup$ By the way, could anyone tell me how to draw commutative diagram in my question and how to add symbols above an arrow? Thanks very much. $\endgroup$
    – Hang
    Commented Jun 27, 2016 at 15:44
  • $\begingroup$ Use the long exact sequence of the pair $(CX,X)$, the fact that for maps of pairs the aforementioned exact sequence is natural, and the fact that the isomorphism (for CW pairs) $H^*(A,B) \cong H^*(A/B)$ is natural. $\endgroup$
    – user98602
    Commented Jun 27, 2016 at 15:51
  • $\begingroup$ @MikeMiller Thank you. Your method may work. But I may prefer to an answer without using CW-complexes. $\endgroup$
    – Hang
    Commented Jun 27, 2016 at 15:54
  • $\begingroup$ Ok, use Mayer-Vietoris on $C^+X \cup C^-X = \Sigma X$. The Mayer-Vietoris sequence is natural for maps that preserve the decomposition. $\endgroup$
    – user98602
    Commented Jun 27, 2016 at 16:22
  • $\begingroup$ @MikeMiller Well, what I want to do is answering this question only by those axioms for homology theory. In my opinion, Mayer-Vietoris argument here is a combination of Homotopy axiom and Excision axiom. And are the two axioms natural? $\endgroup$
    – Hang
    Commented Jun 27, 2016 at 16:36

1 Answer 1


All the constructions that you used to define the isomorphism are natural/functorial:

  • Given a map $X \to Y$, you have a natural map that respect inclusions, which gives a starting point for all the applications of naturality to come: $$(\Sigma X, C_+ X, C_- X, X \times \{0\}) \to (\Sigma Y, C_+ Y, C_- Y, Y \times \{0\});$$
  • The long exact sequence of a pair is natural, hence by using the natural map $(\Sigma X, C_+ X) \to (\Sigma Y, C_+ Y)$, the isomorphism $\tilde{H}_i(\Sigma X) \to \tilde{H}_i(\Sigma X, C_+ X)$ is natural in $X$;
  • Excision is natural, hence the excision isomorphism $\tilde{H}_i(C_- X, X) \to \tilde{H}_i(\Sigma X, C_+ X)$ is natural in $X$;
  • Finally the long exact sequence is still natural, hence the isomorphism $\tilde{H}_i(C_- X, X) \to \tilde{H}_{i-1}(X)$ is natural in $X$.

In conclusion, each subsquare in the following diagram is commutative, hence the "outer" rectangle (by inverting the horizontal arrows that go the wrong way, the composite of the whole thing is the suspension isomorphism) is commutative: $$\require{AMScd} \begin{CD} \tilde{H}_i(\Sigma X) @>{\cong}>> \tilde{H}_i(\Sigma X, C_+ X) @<{\cong}<< \tilde{H}_i(C_- X, X) @>{\cong}>> \tilde{H}_{i-1}(X) \\ @VVV @VVV @VVV @VVV \\ \tilde{H}_i(\Sigma Y) @>{\cong}>> \tilde{H}_i(\Sigma Y, C_+ Y) @<{\cong}<< \tilde{H}_i(C_- Y, Y) @>{\cong}>> \tilde{H}_{i-1}(Y) \end{CD}$$

tl;dr The composition of two functors is a functor.

  • $\begingroup$ Thank you very much for your well-organized answer! I believe I have understood most part of this problem. But, I am sorry that I fail to understand some details. As you said, the long exact sequence of a pair is natural, but I am not sure whether the reduced long homology sequence is natural as well. For example, in general the isomorphism $H_0(X) \to \tilde H_0(X) \oplus G$ is not natural, where $G$ is the coefficient group. $\endgroup$
    – Hang
    Commented Jul 2, 2016 at 12:22
  • $\begingroup$ And, we may not apply excision axiom directly here, instead, we probably have to apply first excision axiom and then homotopy axiom for a deformation retract. So, in order to achieve our goal, is homotopy axiom natural in any sense? $\endgroup$
    – Hang
    Commented Jul 2, 2016 at 12:29
  • $\begingroup$ @Henry Reduced homology is not defined like that, though. It's the homology of the chain complex $\dots \to C_1(X) \to C_0(X) \xrightarrow{\epsilon} G \to 0$, and this is clearly functorial in $X$ (the induced map $C_0(X) \to C_0(Y)$ commutes with the augmentation $\epsilon$). As for homotopy, you need to find a deformation retract that's natural in $X$, which is doable. $\endgroup$ Commented Jul 2, 2016 at 12:32
  • $\begingroup$ Perhaps I am too inquisitive. Can we explicitly find such a deformation retract that is natural? Or any idea of this? $\endgroup$
    – Hang
    Commented Jul 2, 2016 at 15:20
  • 2
    $\begingroup$ I'm a bit late on this, but how does one show that excision is natural? Any references? $\endgroup$
    – Oscar
    Commented May 23, 2021 at 12:51

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