# Regarding continuity of a certain function

This is a part of hatcher's question 2.2.26. For a pair $$(X,A)$$, let $$X\cup CA$$ be $$X$$ with a cone on A attached.
a)show that $$X$$ is a retract of $$X\cup CA$$ iff $$A$$ is contractible in $$X$$.
b)Show that if $$A$$ is contractible in $$X$$ then $$H_n(X,A)=\tilde{H}_n(X)\bigoplus\tilde{H}_{n-1}(A)$$.
Proof:Let $$X$$ be a retract of $$X\cup CA$$. Let $$r$$ denote this retraction. We know $$CA$$ deformation retracts to its tip. Let us call this deformation retraction $$f_t$$. Let us denote $$r|CA = r'$$. consider $$r'(f_t)$$. Consider $$r'(f_t)|A$$. It is evident $$r'(f_0)=i:A\rightarrow X$$ whereas $$r'(f_1)= p\in X$$. Thus $$A$$ is indeed contractible in $$X$$.
Conversely let there be a homotopy $$f_t:A\rightarrow X$$ be such that $$f_0=i:A\rightarrow X$$ and $$f_1$$ is a constant map. Construct a map $$F:CA\rightarrow X$$ as $$F([x,t])=f_t(x)$$. Now we construct a retraction as follows ; Let $$r|X=1$$ and $$r|CA=F$$ this is a retraction onto $$X$$ and is moreover continuous because it agrees on $$A$$.Thus we have a retraction.\ b) Note $$X\cup CA/X$$ is the suspension $$SA$$ of $$A$$. Since $$A$$ is contractible in $$X$$,$$X\cup CA$$ retracts onto $$X$$. This means by the splitting lemma $$\tilde{H}_n(X\cup CA)=\tilde{H}_n(X)\bigoplus \tilde{H}_n(X\cup CA/X)=\tilde{H}_n(X)\bigoplus \tilde{H}_n(SA)=\tilde{H}_n(X)\bigoplus \tilde{H}_{n-1}(A)$$. Also note $$\tilde{H}_n(X\cup CA)=H_n(X\cup CA)=H_n(X\cup CA-{p},CA-{p})=H_n(X,A)$$ where $$p$$ is the tip of the cone. The first isomorphism comes from the exact sequence of the pair while the second from excision, and the third from the deformation retraction of $$CA-{p}$$ onto $$A$$.Thus we get the required result $$H_n(X,A)=\tilde{H}_n(X)\bigoplus\tilde{H}_{n-1}(A)$$.

So my question is in my latter arguement regarding the converse(contractible means a retraction exists) I assume that $$A$$ is closed in $$X$$. Is there a way to extend this to an arbitrary pair (X,A) or do i have to redo my argument altogether.

You do not need to assume that $$A$$ is closed in $$X$$. The space $$X \cup CA$$ is formally defined as follows.
Let $$p : A \times I \to CA$$ denote the quotient map which collapses $$A \times \{1\}$$ to a point. Then $$A' = p(A \times \{0\})$$ is a homeomorphic copy of $$A$$ (the base of $$CA$$). In fact, $$i_0 : A \to CA, i_0(a) = p(a,0) =[a,0]$$, is an embedding with image $$A'$$. Set $$i' : A' \to X, i'([a,0]) = a$$. This is an embedding corresponding to the inclusion $$A \to X$$ under the canonical identification of $$A$$ with $$A'$$ via $$i_0$$. Now $$X \cup CA$$ is a lax notation for the adjunction space $$X \cup_{i'} CA$$, i.e. the quotient of the disjoint sum $$X \sqcup CA$$ obtained by identifying $$[a,0] \in CA$$ with $$a \in X$$. Let $$q : X \sqcup CA \to X \cup CA$$ denote the quotient map.
The map $$R : X \sqcup CA \to X, R(x) = x, R([a,t]) = F([a,t])$$, is continuous. It has the property $$R([a,0]) = F([a,0]) = f_0(a) = i(a) = a = R(a)$$, hence it induces a continuous map $$r : X \cup CA \to X$$ such that $$r \circ q = R$$. Here we do not need that $$A$$ is closed.