It is well known that under some assumptions on a pair $(X,A)$ of a topological space and a subspace we have $H_n(X,A)\simeq \tilde{H}_n(X/A)$. Such an assumption can be for example that $A$ is closed and has an open neighbourhood that deformation retracts onto it, a.k.a. a good pair.

My question is whether there is a natural isomorphism? More precisely:

Consider the category of good pairs $(X,A)$. Are $H_n(X,A)$ and $\tilde{H}_n(X/A)$ functors naturally isomorphic?

I went through Allen Hatcher's "Algebraic Topology", Proposition 2.22. which shows that the quotient map $q:(X,A)\to (X/A, A/A)$ induces an isomorphism $q_n:H_n(X,A)\to H_n(X/A, A/A)$. And I think that this isomorphism is natural, which can be seen from the proof (although I'm not 100% sure). Hatcher finishes the proof with $H_n(X/A,A/A)\simeq \tilde{H}_n(X/A)$ which is fine.

But this brings another question: is the last isomorphism natural? Again, Hatcher mentions in Example 2.18 that this isomorphism arises from the long exact sequence of reduced homology. This again is fine. But does this imply that the isomorphism is natural? I couldn't verify that.

  • $\begingroup$ Yes, the isomorphism is natural. This follows essentially from uniqueness of inverses. $\endgroup$
    – Tyrone
    Commented Jul 4, 2019 at 10:33
  • $\begingroup$ @Tyrone I'm confused, how unique inverses are related to naturality? All isomorphisms (in any category) have unique inverses. $\endgroup$
    – freakish
    Commented Jul 4, 2019 at 11:27
  • $\begingroup$ That's the point. If you have a natural transformation $\alpha$, each of whose components is an isomorphism, then the component-wise defined $\alpha^{-1}$ is also a natural transformation. The issue for you seems to be that you have two different natural isomorphisms pointing in opposite directions. Now you don't. $\endgroup$
    – Tyrone
    Commented Jul 4, 2019 at 12:57
  • $\begingroup$ @Tyrone no, the issue is why those transformations are natural. Sorry for the confusion. $\endgroup$
    – freakish
    Commented Jul 4, 2019 at 13:06

1 Answer 1


Firstly, $q_n:H_n\left(X,A\right)\to H_n\left(X/A,A/A\right)$ is natural and this sort of comes from definition of $H_n\left(X/A,A/A\right)$ as a functor. How would you define this as a functor from pairs of spaces to abelian groups?

Given $f:\left(X,A\right)\to\left(Y,B\right)$, you have the following commutative diagram.

$$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} (X,A) & \ra{q_1} & (X/A,A/A) \\ \da{f} & & \da{\overline{f}} \\ (Y,B) & \ras{q_2} & (Y/B,B/B) \\ \end{array} $$

where $\overline{f}$ is just $q_2\circ f$ passed to the quotient since it maps all of $A$ to $B/B$. Therefore you can have the functor $H_n\left(X/A,A/A\right)$ map $f$ to $\overline{f}_n$. But by functoriality of relative homology groups, the above diagram means that

$$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} H_n(X,A) & \ra{{q_1}_n} & H_n(X/A,A/A) \\ \da{f_n} & & \da{\overline{f}_n} \\ H_n(Y,B) & \ras{{q_2}_n} & H_n(Y/B,B/B) \\ \end{array} $$

also commutes, so $q_n$ is a natural transformation, and, in particular, it is a natural isomorphism if you restrict your category to good pairs.

Finally, the isomorphism $H_n(X/A,A/A)\cong \tilde{H}_n(X/A)$ is also natural since the long exact sequence is natural (i.e. each map is natural).

I hope this helps.

  • $\begingroup$ Beautiful! That's the explanation I've been looking for. $\endgroup$
    – freakish
    Commented Jul 4, 2019 at 11:16
  • $\begingroup$ @freakish Thank you! $\endgroup$ Commented Jul 4, 2019 at 16:25

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