Proposition 6.10 of Bredon-Algebraic Topology

Question

In this context Bredon considers an homology theory with integer coefficient and he proves the following:

I don't understand how he determines the maps $f_\ast:(a,b)\longmapsto (b,a)$ and $S^0\longrightarrow P$.

Answer 1

If you use singular homology, you have the $0$-simplex: (Recall that $\Delta^0 = P$).

• $\sigma_x : \Delta^0 \rightarrow S^0$ (sent the point to $x$).

• $\sigma_y : \Delta^0 \rightarrow S^0$ (sent the point to $y$).

• $\sigma : \Delta^0 \rightarrow P$ just the identity map.

And $C_0(S^0) = \{ m\sigma_x + n\sigma_y : m,n \in \mathbb{Z}\}$, $C_0(P) = \{m\sigma : m \in \mathbb{Z})$,

The map $S^0 \rightarrow P$ induces a map at the level of chain complexes sending $ m\sigma_x + n\sigma_y \mapsto (m+n)\sigma$

because the composition $\Delta_0 \xrightarrow{\sigma_x} S^0 \rightarrow P$ is equal to $\sigma$, and the same for $\sigma_y$.

You can use a similar argument just over the generators of $C_0(S^0)$ to conclude that $f_*$ is the swapping map induced in homology.

Answer 2

This is a nice question, since the details can be a bit blurry. So, let's call $X_{-1}=\{-1\}$ and $X_1=\{1\}$. We then have a canonical map $c:S^0 \to X_{-1}\oplus X_1$ (disjoint union). By the additivity axiom, we have that $$\iota_{-1*}\oplus \iota_{1*}: H_0(X_{-1}) \oplus H_0(X_1)\to H_0(X_{-1}\oplus X_1) $$ is an isomorphism. Consider now $\eta: X_{-1} \oplus X_1 \to S^0$ the obvious map. Being a homeomorphism, we have that $\eta_*$ is an isomorphism. We then have the commutative diagram (with $f_*''$ and $f_*'$ defined in order for the diagram to commute). $\require{AMScd}$ \begin{CD} H_0(X_{-1}) \oplus H_0(X_1) @>\iota_{-1*}\oplus \iota_{1*}>> H_0(X_{-1}\oplus X_1) @>\eta_*>> H_0(S^0)\\ @V f_*'' V V @VV f_*' V @V f_* VV \\ H_0(X_{-1}) \oplus H_0(X_1) @>\iota_{-1*}\oplus \iota_{1*}>> H_0(X_{-1}\oplus X_1) @>\eta_*>> H_0(S^0) \end{CD} When he says that "$f_*$ becomes, in the direct sum" etc, he is almost referring to $f_*''$. Why "almost"? Because being an interchange does not even make sense yet. Now, we have a unique map $m_{-1}: X_{-1} \to X_1$ (and an analogous $m_1$). This is the "unique maps between them" which identify the homology of all point spaces as Bredon says. Being a homeomorphism, this induces an isomorphism $m_{-1*}$, which fills the diagram as follows $\require{AMScd}$ \begin{CD} H_0(X_{-1}) \oplus H_0(X_{-1}) @> Id \oplus m_{-1} >> H_0(X_{-1}) \oplus H_0(X_1) @>\iota_{-1*}\oplus \iota_{1*}>> H_0(X_{-1}\oplus X_1) @>\eta_*>> H_0(S^0)\\ @V f_*''' V V @VV f_*'' V @V f_*' VV @V f_* VV \\ H_0(X_{-1}) \oplus H_0(X_{-1}) @> Id \oplus m_{-1} >> H_0(X_{-1}) \oplus H_0(X_1) @>\iota_{-1*}\oplus \iota_{1*}>> H_0(X_{-1}\oplus X_1) @>\eta_*>> H_0(S^0). \end{CD} It is $f_*'''$ he is talking about. Now, note that $m_1 \circ p_1 \circ \eta^{-1} \circ f \circ \eta \circ \iota_{-1}=Id_{X_{-1}}$, where $p_i:X_{-1} \oplus X_1 \to X_i$ is the obvious map. Now, the work is almost done. Functoriality and some more additions to the diagram will yield the result (I don't know how to make diagonal arrows here and the answer is already very big, so I'll end it here).