Isomorphism not natural in $X$?

Solution 1:

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 $\widetilde{(f_1)}_* = \widetilde{(f_2)}_* = 0$.