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Let $p_0 : A \to X_0 $ and $p_1 : A \to X_1$ be two maps. I am confused about what does it mean to say that '$X_0$ and $X_1$ are homotopy equivalent under $A$'. Which of the following statements is this equivalent to :

1.) $\exists f: X_0 \to X_1, g: X_1 \to X_0$ such that $f \circ g \sim_{p_1} id_{X_1}$ and $g \circ f \sim_{p_0} id_{X_0}$

or

2.) $\exists f,g$ as in condition (1) which are further required to be maps under $A$ i.e. $f \circ p_0 =p_1$ and $g \circ p_1 = p_0$.

The confusion arises because problem 8 on page 207 of Algebraic Topology by Tammo tom Dieck asks to prove that homotopic attaching maps give rise to homotopy equivalent (under the lower dimension skeleton) CW-complexes. This does not seem to be true if one understands it to mean the stronger condition (2) while my intuition says that the correct meaning of the terminology is (2). Hence the confusion

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  • $\begingroup$ If by $gf\sim_{p_0}id_{X_0}$ you mean that the homotopy is "under A", that is, $H:X_0\times I\to X_0$ is a map with $H(x,0)=x,\ H(x,1)=gf(x)$ and $H(p_0(a),t)=p_0(a)$, then the second meaning should be correct. $\endgroup$ – Stefan Hamcke Apr 29 '15 at 15:14
  • $\begingroup$ @StefanHamcke If we accept the second meaning, does the following hold : Let $L_0, L_1$ be obtained from $K$ by attaching maps $\phi_0, \phi_1$ respectively. If the attaching maps are homotopic, it is true that $L_0, L_1$ will be homotopy equivalent under $K$? $\endgroup$ – user90041 Apr 29 '15 at 17:21
  • $\begingroup$ So if I understand correctly, $\phi_0,\phi_1:A\to K$ are homotopic maps from some subspace $A\subseteq X$, and $L_i=X\cup_{\phi_i}K$. If the pair $(X,A)$ is cofibered, then $L_0$ is homotopy equivalent to $L_1$ under $K$. $\endgroup$ – Stefan Hamcke Apr 29 '15 at 20:59
  • $\begingroup$ @StefanHamcke Exactly. That is what we need to show. But I am confused because it does not seem to be true in this example. Let $A= \amalg S^{1}, X= \amalg D^2, K=\mathbb{R}$. Let the attaching maps be the constant maps to to distinct integer points of $\mathbb{R}$. Then the resulting spaces are homotopy equivalent (in fact homeomorphic) but they do not seem to be homotopy equivalent under $\mathbb{R}$. Please advise if I am making some mistake. Thanks (We can assume that the space being attached is an n-cell and that the original space is a CW-complex.) $\endgroup$ – user90041 Apr 30 '15 at 5:13
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The result is

Let $(X,A)$ be cofibered and assume $f_0,f_1:A\to Y$ are maps which are homotopic by $H_t:A\to Y$. Then $Z_0=X\cup_{f_0}Y$ and $Z_1=X\cup_{f_1}Y$ are homotopy equivalent rel $Y$.
Proof: Let $Z=X\times I\cup_H Y$ be the space obtained by gluing $X×I$ along the homotopy on $A×I$ to $Y$. This space contains $Z_0$ as the subspace $(X×\{0\}\cup A×I)\cup_H Y$, and likewise $Z_1$. If $q:X×I\sqcup Y\to Z$ is the quotient map, then the product map $q×\mathbf 1_I:X×I×I\sqcup Y×I\to Z×I$ is a quotient map as well since $I$ is locally compact. Therefore a homotopy on $Z$ is given by compatible homotopies $X×I\to Z$ and $Y\to Z$. On $X×I$ let the homotopy send $(x,s,t)$ to $d(x,s,t)$, where $d:X×I×I\to X×I$ is the homotopy retracting $X×I×I$ to $X×\{0\}\cup A×I$. On $Y$ let the homotopy be independent on $t$. The resulting homotopy on $Z$ retracts $Z$ onto $Z_0$. Similarly, there's a homotopy retracting $Z$ onto $Z_1$. Since the maps $Z_0\hookrightarrow Z$ and $Z\to Z_1$ are homotopy equivalences rel $Y$, so is their composition.

Here is an example, which is somewhat trivial since both adjunction spaces are deformation retracting to $Y$, but it serve as an aid to see how the homotopy looks like.
Let $X=Y=I$ and $A=\{0\}$ and let $f_i(0)=i$ with the homotopy $H_s(0)=s$. We can think of $Z_0$ as $I×\{0\}\cup\{0\}×I$ and of $Z_1$ as $I×\{1\}\cup\{0\}×I$. If $g:Z_0\to Z_1$ is the map resulting from embedding $Z_0$ into $Z$ and then composing with the deformation retraction onto $Z_1$, and similarly for $h:Z_1\to Z_0$, then $hg$ sends $[0,1/2]×\{0\}$ and $[1/2,3/4]$ convex-linearly to $\{0\}×I$, and $[3/4,1]×\{0\}$ to $I×\{0\}$ with $hg(0,0)=(0,0),\ hg(1/2,0)=(0,1),\ hg(3/4,0)=(0,0)$, and $hg(1,0)=(1,0)$, and the homotopy draws the loop $g([0,3/4])$ into $I×\{0\}$.

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  • $\begingroup$ Thanks so much, I think I am understanding the proof now. However in the example, I am having some confusion about the definition of the map $g$. There appear to be some minor typos which are probably causing it. I am not sure how $g$ is defined on $[3/4,1] \times 0$. Also the RHS of $g(1,0)=(1,0)$ does not belong to $Z_1$. Could you please edit it ? $\endgroup$ – user90041 May 1 '15 at 6:36
  • $\begingroup$ @user90041: Sorry, I corrected it. The map $g$ sends the lower half of $I×\{0\}$ onto $\{0\}×I$ and the upper half onto $I×\{1\}$. What I wrote as $g$ in my answer should actually be the composition of the two homotopy equivalences $g$ and $h$. $\endgroup$ – Stefan Hamcke May 1 '15 at 15:11

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