Spectra and long exact sequences

Question

Suppose $X,Y,Z$ are finite CW complexes and let $f \colon X \to Y$ be a fibration with fiber $F$. Is there now a long exact sequence associated to their suspension spectra in the following sense $$ \ldots \to [\Sigma^k Z,\Sigma^k F] \to[\Sigma^k Z,\Sigma^k X] \to [\Sigma^k Z,\Sigma^k Y] \to [\Sigma^{k-1} Z,\Sigma^{k-1} F] \to \ldots\ $$ If not, which long exact sequence can on associate with $Z$ and the fibration $f$?

Answer 1

Given a map $f:X\rightarrow Y$ between CW complexes you form its cofiber $C_f$ to get a cofibration sequence starting

$$X\xrightarrow fY\xrightarrow iC_f\xrightarrow q\Sigma X\xrightarrow{\Sigma f}\Sigma Y\rightarrow\dots.$$

This sequence is stable under suspension, in that you can continue it to the right indefinitely as

$$\dots\Sigma^{k-1}C_f\xrightarrow{\Sigma^{k-1}q} \Sigma^kX\xrightarrow{\Sigma^k f}\Sigma^kY\xrightarrow{\Sigma^k i}\Sigma^kC_f\xrightarrow{\Sigma^k q}\Sigma^{k+1} X\rightarrow\dots$$

In this way, for any space $Z$ you get a contravariant exact sequence

$$\dots\leftarrow[\Sigma^{k-1}C_f,Z]\xleftarrow{\Sigma^{k-1}q}[\Sigma^kX,Z]\xleftarrow{\Sigma^kf^*} [\Sigma^kY,Z]\xleftarrow{\Sigma^ki^*}[\Sigma^kC_f,Z]\xleftarrow{\Sigma^kq^*}[\Sigma^{k+1} X,Z]\xleftarrow{\Sigma^{k+1}f^*}\dots\qquad(*)$$

which is an exact sequence of abelian groups for $k>\geq 2$, of groups when $k=1$ and of sets when $k=0$.

You can also form a covariant sequence

$$\dots\rightarrow[Z,\Sigma^kX]\xrightarrow{\Sigma^k f_*}[Z,\Sigma^kY]\xrightarrow{\Sigma^k i_*}[Z,\Sigma^kC_f]\xrightarrow{\Sigma^k q_*}[Z,\Sigma^{k+1}X]\xrightarrow{\Sigma^{k+1} f_*}[Z,\Sigma^{k+1}Y]\rightarrow\dots\qquad(**)$$

which will not in general be exact. However there is something that can be said. It is certainly the case that that the composition of two consecutive homomorphisms in this sequence is trivial, and the deeper theorem actually states that this sequence is exact in a certain range of dimensions. That is, if the dimension of $Z$ is 'small' relative to $k$, then the sequence will be exact.

Let me elaborate a little. Recall the map $q$ which I introduced in the very first cofibration sequence, and form its homotopy fibre $F_q$ to get a fibration sequence

$$\dots\rightarrow F_q\xrightarrow{r}Y\xrightarrow{q} C_f$$

Now there is a canonical nullhomotopy of the composite $q\circ f$ that aries from the fact that these two maps appear consecutively in a cofibration sequence. Then by the definition of a fibration sequence this means that you get an induced map

$$\theta:X\rightarrow F_q$$

lifting $f$, and it turns out that this map is an $n$-equivalence, where $n$ is a number that depends on the connectivity of $X$ and $Y$ and the homotopical behaviour of $f$ in a way you can calculate.

As the connectivities of $X$ and $Y$ increase, the integer $n$ increases, and the homotopy fibre $F_q$ looks more and more like $X$. In particular, if we replace $q$ by $\Sigma^kq$, form the homotopy fibre $F_{\Sigma^k q}$ and the comparison map

$$\theta_k:\Sigma^kX\rightarrow F_{\Sigma^kq}$$

then this map will be an $(n+2k)$-equivalence. So in particular, as $k$ increases $\Sigma^kX$ looks more and more like the homotopy fibre of $\Sigma^kq:\Sigma^kY\rightarrow\Sigma^kC_f$. Therefore, if $Z$ is a finite dimensional complex there will be a large $k$ for which the covariant sequence $(**)$ I introduced above will be exact.

Now it is not true that $F_{\Sigma^kq}$ is equivalent to the suspenion $\Sigma F_{\Sigma^{k-1}q}$. Indeed, in general it will not even be a suspension. What we have succeeded in doing is only approximating it by a suspension. The point is that as $k$ increases the approximation gets better and better, to the point at which we are dealing with spectra.

In the ideal limit at which $k$ is taken all the way to infinity we move to the realm of spectra. At this point there is no difference between cofiber and fibre sequences, and this is more or less the definition of stability. When dealing with spectra, both the contravariant sequence $(*)$ and the covariant sequence $(**)$ are exact. Just note that there are defined with the cofibre of $f$ rather than its fibre.