# Isomorphism not natural in $X$

algebraic-topologygeneral-topologyhomology-cohomologynatural-transformationssolution-verification

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I am working on the following task:

Let $$\mathcal{H}_*$$ be a homology theory and let $$X \neq \emptyset$$ be a space. Construct an isomorphism $$\mathcal{H}_n(X) \cong \widetilde{\mathcal{H}}_n(X) \oplus \mathcal{H}_n(*)$$, where $$\widetilde{\mathcal{H}}_*$$ is reduced homology theory and $$*$$ is a point. Can this isomorphism be chosen natural in $$X$$?

Edit: As was pointed out in the comments, by "homology theory $$\mathcal{H}_*$$" I mean a generalized homology theory satisfying the Eilenberg-Steenrod axioms, but not necessarily the dimension axiom. I denote by $$H_*$$ simplicial homology.

I am able to show the first part, but I'm unsure about naturality: assume that the isomorphism were natural. Then for any spaces $$X,Y$$ and map $$f : X \to Y$$ there should exist a unique ("natural") isomorphism $$t : \mathcal{H}_n(-) \to \widetilde{\mathcal{H}}_n(-) \oplus \mathcal{H}_n(*)$$ such that the following diagram commutes:

$$\begin{CD} \mathcal{H}_n(X) @>f_*>> \mathcal{H}_n(Y)\\ @Vt_XVV @VVt_YV \\ \widetilde{\mathcal{H}}_n(X) \oplus \mathcal{H}_n(*) @>f_*\oplus \text{id}>> \widetilde{\mathcal{H}}_n(Y) \oplus \mathcal{H}_n(*) \end{CD}$$

However, if we choose $$X=Y=S^1$$, $$f=\text{id}$$ and $$\mathcal{H}_* = H_*$$ as simplicial homology, so that we can assume the dimension axiom, then we get for $$n=1$$ that $$H_1(S^1) \cong \mathbb{Z}$$, $$H_1(*) \cong 0$$ and $$f_*=\text{id}_\mathbb{Z}$$. So we get the following diagram:

$$\begin{CD} \mathbb{Z} @>\text{id}_\mathbb{Z}>> \mathbb{Z}\\ @Vt_{S^1}VV @VVt_{S^1}V \\ \mathbb{Z} \oplus 0 @>\text{id}_\mathbb{Z}\oplus 0>> \mathbb{Z} \oplus 0 \end{CD}$$

But now we see that we could choose $$t_{S^1}$$ as the isomorphism $$n \mapsto (n,0)$$ or $$n \mapsto (-n,0)$$ and the above diagram would commute in both cases. Therefore, since we have a choice for $$t_{S^1}$$, the isomorphism is not natural in $$X$$.

Is my reasoning correct?

Let us first understand the isomorphism $$\mathcal{H}_n(X) \cong \widetilde{\mathcal{H}}_n(X) \oplus \mathcal{H}_n(*)$$. You define reduced homology groups by $$\widetilde{\mathcal{H}}_n(X) = \ker (c_* : \mathcal{H}_n(X) \to \mathcal H_n(*)) \tag{R}$$ where $$c : X\to *$$ denotes the unique map to the one-point space $$*$$.

Clearly each $$f : X \to Y$$ has the property $$f_*(\widetilde{\mathcal{H}}_n(X)) \subset \widetilde{\mathcal{H}}_n(Y)$$, i.e. we get an induced homomorphism $$\widetilde f_* : \widetilde{\mathcal{H}}_n(X) \to \widetilde{\mathcal{H}}_n(Y)$$ which satisfies $$i_Y \widetilde f_* = f_* i_X$$. It is easy to see that this construction gives us functors $$\widetilde{\mathcal{H}}_n$$.

By definition $$\text{(R)}$$ we get the short exact sequence $$0 \to \widetilde{\mathcal{H}}_n(X) \stackrel{i}{\hookrightarrow} \mathcal{H}_n(X) \stackrel{c_*}{\to} \mathcal H_n(*) \to 0 \tag{E}$$ Note that $$c_*$$ is an epimorphism because it has a right inverse. In fact, each map $$j : * \to X$$ has the property $$c_* j_* = id$$. The latter also shows that $$\text{(E)}$$ is a split short exact sequence.

Therefore each $$j : * \to X$$ induces an isomorphism $$j_\# : \widetilde{\mathcal{H}}_n(X) \oplus \mathcal H_n(*) \to \mathcal H_n(X), j_\#(a,b) = i(a) + j_*(b) \tag{S}$$ More generally we could take any right inverse $$\iota : \mathcal H_n(*) \to \mathcal H_n(X)$$ of $$c_*$$ to produce an isomorphism $$\iota_\# : \widetilde{\mathcal{H}}_n(X) \oplus \mathcal H_n(*) \to \mathcal H_n(X), \iota_\#(a,b) = i(a) + \iota(b) \tag{S'}$$ In general not all of these $$\iota$$ will have the form $$\iota = j_*$$ for some $$j$$. Anyway, the assocation $$\iota \mapsto i_\#$$ described in $$\text{(S')}$$ establishes a bijection from the set of right inverses $$\iota$$ of $$c_*$$ to the set of isomorphisms $$\widetilde{\mathcal{H}}_n(X) \oplus \mathcal H_n(*) \to \mathcal H_n(X)$$.

For a path-connected $$X$$ all maps $$j: * \to X$$ are homotopic, i.e. $$j_*$$ does not depend on the choice of $$j$$. We therefore get a canical isomorphism $$J = J_X : \widetilde{\mathcal{H}}_n(X) \oplus \mathcal H_n(*) \to \mathcal H_n(X)$$ which has the form $$J =j_\#$$ for any choice of $$j$$.

On the full subcategory of path-connected spaces the above $$J_X$$ form a natural isomorphism. This follows from

Lemma. Let $$Y$$ be path connected, $$f : X \to Y$$ and $$j_X : * \to X$$. Then $$J_Y (\widetilde f_* \oplus id) = f_*(j_X)_\#$$.

Proof. We have $$J_Y = (j_Y)_\#$$ with $$j_Y = f j_X$$ and we get $$J_Y(\widetilde f_* \oplus id)(a,b) = (j_Y)_\# (\widetilde f_* \oplus id)(a,b) = (j_Y)_\#(\widetilde f_*(a),b) = i_Y \widetilde f_*(a) + (j_Y)_*(b) \\= f_*i_X(a) + (fj_X)_*(b) = f_*i_X(a) + f_*(j_X)_*(b) = f_*(i_X(a) + (j_X)_*(b)) = f_*(j_X)_\#(a,b) .$$

For the general case we have

Theorem. There exists a natural isomorphism $$t_X : \widetilde{\mathcal{H}}_n(X) \oplus \mathcal H_n(*) \to \mathcal H_n(X)$$ on the category of all toplogical spaces if and only if $$\mathcal H_n(*) = 0$$. NB: $$\mathcal H_n(*) = 0$$ does not imply that $$\mathcal H_n(X) = 0$$ for all $$X$$.

Proof. If $$\mathcal H_n(*) = 0$$, then $$\widetilde{\mathcal{H}}_n(X) = \mathcal H_n(X)$$. We can take $$t_X = id$$.

Now let $$\mathcal{H}_n(*) \ne 0$$. Then $$\widetilde{\mathcal{H}}_n(*) = 0$$. Consider $$Y = \{1,2\}$$ with the discrete topology. We have two maps $$f_i : * \to Y$$ given by $$f_i(*) = i$$. The excision axiom gives us an isomorphism $$\phi : \mathcal{H}_n(Y) \to \mathcal{H}_n(*) \oplus \mathcal{H}_n(*)$$ which has the property that $$\phi (f_i)_*$$ is the inclusion in the $$i$$-th summand. Thus $$(f_1)_* \ne (f_2)_*$$. If there would exist a natural isomorphism $$t_X$$, we would get two commutative diagrams

$$\begin{CD} \mathcal{H}_n(*) @>(f_i)_*>> \mathcal{H}_n(Y)\\ @Vt_*VV @VVt_YV \\ 0 \oplus \mathcal{H}_n(*) @>\widetilde{(f_i)}_*\oplus \text{id}>> \widetilde{\mathcal{H}}_n(Y) \oplus \mathcal{H}_n(*) \end{CD}$$ But this is impossible because $$t_Y(f_1)_* \ne t_Y(f_2)_*$$, but $$\widetilde{(f_1)}_* = \widetilde{(f_2)}_* = 0$$.