Prove for cohomology theory $E^\bullet$, there is an isomorphism $E^\bullet(X,A) \stackrel{\simeq}{\longrightarrow} E^\bullet(X \cup Cone(A), \ast)$ This is a question from the proof for Lemma 2.8 of (https://ncatlab.org/nlab/show/generalized+%28Eilenberg-Steenrod%29+cohomology).
Given an cohomology theory $E^\bullet$ and $A \hookrightarrow X$, the lemma claims that $E^\bullet(X,A) \stackrel{\simeq}{\longrightarrow} E^\bullet(X \cup Cone(A), \ast)$, where $Cone(A)$ is the reduced cone of $A$.
(I think $Cone(A)$ in this lemma is reduced, because it is then used to prove proposition 3.2, where an unreduced cohomology theory can give a reduced cohomology theory.)
According to the proof of the lemma, we define $U = (Cone(A)-A \times \{0\}) \hookrightarrow Cone(A)$, and then we try to use excision axiom to show $E^\bullet (X\cup Cone(A), Cone(A)) \stackrel{\simeq}{\longrightarrow} E^\bullet (X\cup Cone(A)-U, Cone(A)-U)$.
The question that confuses me is: The excision axiom requires $\overline{U} \subset Int(Cone(A))$, which seems quite strange and not always satisfied when $Cone(A)$ is reduced. For example, when the basepoint $*$ is on the boundary of $A$, $\overline{U}$ can contain $*$ while $*$ is not in $Int(Cone(A))$. So how to deal with it？
Thanks in advance!
 A: The lemma deals with an unreduced  cohomology theory. No basepoints are involved on this case, but unfortunately the lemma does not make precise what is meant by $Cone(A)$ in this case. Actually $X \cup Cone(A)$ denotes the mapping cone of the inclusion $i: A \hookrightarrow X$. Writing $X \cup Cone(A)$ is perhaps a bit sloppy because it suggests that $X \cup Cone(A)$ is an "ordinary union" of $X$ and $Cone(A)$ and does not make explicit that a gluing process in involved here. In fact $X \cup Cone(A)$ is the adjunction space obtained frm $X$ by attaching $Cone(A)$ via the obvious embedding of its base $A \times \{0\}$ into $X$. However, $X \cup Cone(A)$ indeed contains a copy of the unreduced cone $Cone(A)$ as a subspace.
The proof is based on the excision of some $U \subset Cone(A)$ to get an isomorphism
$$\iota^* : E^n(X \cup Cone(A), Cone(A)) \to E^n((X \cup Cone(A)) \setminus
U, Cone(A)\setminus U) . \tag{1}$$
Unforunately $U = Cone(A) \setminus A \times \{0\}$ does not work because  $\overline U = Cone(A)$ which is in general not contained in $Int(Cone(A))$ (it may only be in some special cases).
Instead we take $U = Cone(A) \setminus A \times [0,1/2]$; then $\overline U \subset Int(Cone(A))$ and we see that $(1)$ is an isomorphism.
The space $(X \cup Cone(A)) \setminus U$ is a copy of the mapping cylinder of the inclusion $i$. Let $r : (X \cup Cone(A)) \setminus U \to X$ be a strong deformation retraction. This gives a homotopy equivalence of pairs $r : ((X \cup Cone(A)) \setminus U, Cone(A) \setminus U) \to (X,A)$. Thus
$$r^* : E^n(((X \cup Cone(A)) \setminus U, Cone(A) \setminus U) \to E^n(X,A) \tag{2}$$
is an isomorphism which shows that we get an isomorphism
$$E^n(X \cup Cone(A), Cone(A)) \to E^n(X,A) .$$
In a final step we get an isomorphism
$$\phi^* : E^n(X \cup Cone(A), Cone(A)) \to E^n(X \cup Cone(A),*) \tag{3]$$
where $*$ is any point of $Cone(A)$, for example its tip. This isomorphism is induced by the inclusion of pairs $\phi : (X \cup Cone(A), *) \to (X \cup Cone(A), Cone(A))$. This is not a homotopy equivalence of pairs, but the "absolute map" $X \cup Cone(A) \to X \cup Cone(A)$ is the identity and the "restricted map" $* \to Cone(A)$ is a homotoppy equivalence. The long exact sequences of the pairs $(X \cup Cone(A), *)$ and $(X \cup Cone(A), Cone(A))$ are connected by $\phi$ and its restrictions, hence the five lemma shows that $(3)$ is an isomorphism.
Let me finally remark that the I do not recommend nLab as a good source for generalized cohomology. Better consult the references in the nLab-article.
