Showing that a sequence of a cofibration is exact.

algebraic-topologycofibrationsexact-sequencehomotopy-theory

The question is :

Suppose that $A \rightarrow B$ is a cofibration with cofiber $C.$ Show that for any pointed space $X,$ the sequence $[C,X] \rightarrow [B,X] \rightarrow [A,X]$ is an exact sequence of pointed sets.

My professor answer started with:

"if $A \xrightarrow{\text{i}} B \xrightarrow{\text{q}} B/A$ is a cofibre sequence " but I do not know why he started by this statement, does every cofibre sequence has the inclusion map and the quotient map into $B/A,$ could anyone explain this for me please?

Also, the definition of cofibration in AT does not say the above, may be he is using pg.398 in AT, I do not know.

Best Answer

Terminology some people use is to reserve the term cofibre sequence, or strict cofibre sequence, for something of the form

$$A\xrightarrow{i} B\xrightarrow{q} B/A$$

where $i$ is a cofibration (i.e. has the HEP) and $q$ is the quotient. Sequences of this form have the nice property that strict equality holds for the composition

$$q\circ i=\ast.$$

That this equality holds strictly makes working with these spaces an maps more easy in many applications.

On the other hand, we can turn any map $f:X\rightarrow Y$ into a cofibration by replacing it with the inclusion of $X$ into the mapping cyclinder $M_f$ of $f$. This gives us a strict cofibre sequence

$$X\xrightarrow{i_f} M_f\xrightarrow{q_f} M_f/i_f(X).$$

Here $M_f=Y\sqcup X\times I/[f(x)\sim (x,0)]$, and $i_f$ is the map $x\mapsto (x,1)$. You can check easily that $i_f$ is a cofibration. We can also identify the quotient

$$M_f/i_f(X)=C_f=Y\sqcup CX/[f(x)\sim (x,0)]$$

with the mapping cone $C_f$ of $f$ (here $CX=X\times I/X\times 1)$.

Now there is an obvious inclusion $Y\hookrightarrow M_f$ which turns out to be a deformation retract. The point is that up to identification by homotopy we can write the second sequence above as

$$X\xrightarrow{f}Y\xrightarrow{q} C_f$$

This is what some authors would call a homotopy cofibre sequence. Compared to the above sequences we now have only the relation of null-homotopy

$$q\circ f\simeq \ast$$

rather than the strict equality in the previous cases. Note however, that there is a canonical null-homotopy of this composition (just slide up the cone in $C_f$).

Of course, if we are working at the level of homotopy, then really this is all we need: strict equality is essentially meaningless here. Hence a lot of authors will call both types of sequences above simply cofibre sequences, and not make any distinction between the two, since it doesn't really effect things at the level of homotopy.

So, to answer your question, let us define a cofiber sequence to be a pair, consisting of a sequence

$$X\xrightarrow{f}Y\xrightarrow{g}Z$$

of spaces and maps, together with a choice of null-homotopy $\psi:g\circ f\simeq \ast$, such that the map $\theta_\psi:C_f\rightarrow Z$ induced by $\psi$ is a homotopy equivalence.

This definition encompasses both the previous constructions, and with it not every cofibre sequence is exactly a subspace inclusion followed by a quotient map, but every cofibre sequence is equivalent to one of this form (in a sense you will come to understand more precisely in your lectures).

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