Let $X$ be a topological space and suppose $X_1$ and $X_2$ are spaces obtained by attaching an n-cell to $X$ via homotopic attaching maps. Show that $X_1$ and $X_2$ are homotopy equivalent.


So, my idea is to use the fact that if two spaces are homeomorphic to deformation retracts of a single larger space then they are homotopy equivalent to one another. So, I would need to either show that both $X_1$ and $X_2$ are deformation retracts of some space, or homeomorphic to some deformation retracts of this larger space.

Since n-cells are homeomorphic to the balls $\mathbb{B}^n$, ( with $\mathbb{B}^n$ we are denoting the closed n-ball here) the idea would be to consider the larger adjunction space $X\bigcup_{\phi}\mathbb{B}^n\times I$, where the attaching map $\phi:\mathbb{B}^n\times \{0\}\rightarrow X$ would be given by $\phi(x,0)=f(x)$, where $f:\mathbb{B}^n\rightarrow X$ would be some continuous map.

On the other hand we know that $X_1$ is the space obtained from $X$ by attaching an $n$-cell, so it is homeomorphic to an adjunction space, more specifically: $X\bigcup_{f_1}\mathbb{B}^n$, where $f_1:\partial\mathbb{B}^n\rightarrow X,$ is the corresponding attaching map. Similarly, $X_2$ is homeomorphic to the adjunction space $X\bigcup_{f_2}\mathbb{B}^n$, where $f_2:\partial\mathbb{B}^n\rightarrow X,$ is the corresponding attaching map.

So, I guess, now to conclude the proof, it would suffice to show that $X\bigcup_{f_1}\mathbb{B}^n$ and $X\bigcup_{f_2}\mathbb{B}^n$ are deformation retracts of $Z_f:=X\bigcup_{\phi}\left(\mathbb{B}^n\times I\right).$

So, we need to define a homotopy: $H:Z_f\times I\rightarrow Z_f.$ Before we do that, let $q:X\amalg\left(\mathbb{B}^n\times I\right)\rightarrow Z_f,$ be the quotient map between the given disjoint union of those spaces and $Z_f.$

Now, let us define the homotopy as follows: $$H(q(b,s),t)=q(b,s(1-t)),\hspace{3mm} H(q(x),t)=q(x),$$ where $(b,s)\in \mathbb{B}^n\times I$, and $x\in X.$

Now, $H(q(b,s),0)=q(b,s)$ and $H(q(x),0)=q(x),$ also $H(q(b,s),1)=q(b,0)\in\mathbb{B}^n\times\{0\}$ and $H(q(x),1)=q(x)\in X.$ So, i believe this shows that they are deformation retractions, but i think i am missing something here??

Am I on the right track, and how should i proceed?



2 Answers 2


This should be a comment, but it got too long.

Let $B$ be the ball you want to attach and $S=\partial B$. In your situation, you have a map $\phi:S\times I\to X$ giving an homotopy from $\phi_0$ to $\phi_1$, and you want to compare $X_0=X\cup_{\phi_0}B$ and $X_1=X\cup_{\phi_1}B$.

If we define $\bar\phi:(x,t)\in S\times I\mapsto (\phi(x,t), t)\in X\times I$, we can consider the adjunction space $(X\times I)\cup_{\bar\phi}(B\times I)$, which contains copies of both $X_0$ and $X_1$ on each «end». Can you us this space to construct homopies?

  • $\begingroup$ Why can't I consider the adjunction space $X\bigcup_{\phi}\left(B\times I\right)$? $X_0$ and $X_1$ would still be copies contained in there. $\endgroup$
    – Brandon
    Sep 28, 2012 at 6:49
  • $\begingroup$ Anyone else any hints? $\endgroup$
    – Brandon
    Sep 28, 2012 at 10:28
  • $\begingroup$ Mariano, this problem is Exercise 14 on page 11 of this pdf. I am trying to do Exercises 14 and 15. The author says we can do them without much beyond the definition of homotopy and adjunction spaces, yet everywhere I've seen this question discussed people talk about cofibrations, Homotopy extension property and deformation retracts. Your answer seems to be the most elementary, so could you please tell me a) How to finish your answer off, because I can't see it and b) If Hillman's hints in the pdf give an easy solution? Thank you. $\endgroup$ Jun 18, 2013 at 13:59
  • $\begingroup$ @Brandon The problem with your suggestion is that different copies of $B$ (as $t$ progresses in $I$, so to speak) will be "glued" together when they are mapped to the same $X$. We don't want that. $\endgroup$ Apr 17 at 21:29

I copy the answer from a tex written by me:

Let $(X, A)$ be a relative CW complex and let $f, g : A \to Y$ be maps homotopic equivalent, where $Y$ is a topological space. Then: $$ X \cup_f Y \approx X \cup_g Y \qquad rel Y $$ where by ``$rel Y$'' we mean that the homotopy is constantly the identity on $A$.

proof Let $F : A \times I \to Y$ be a homotopy from $f$ to $g$. Let us consider the space $Z := (X \times I) \cup_F Y$. $Z$ contains $X \cup_f Y$ and $X \cup_g Y$ as subspaces. The strong deformation retraction of $X \times I$ onto $X \times \{0\} \cup A \times I$ gives rise to a retraction from $(X \times I) \cup_F Y$ to $(X \times \{0\} \cup A \times I) \cup_F Y = (X \times \{0\}) \cup_F Y \sim X \cup_f Y$. The same is true between $(X \times I) \cup_F Y$ and $X \cup_g Y$. Those homotopies are constantly the identity on $Y$.

Corollary Consider a relative CW complex $(X, A)$, where $X$ is obtained from $A$ by attaching $n$--cells. Then the homotopy class of $X rel A$ depends only on the homotopy classes of the attaching maps of $(X, A)$.

proof If the attaching maps $f_{\alpha}, g_{\alpha} : S_{\alpha}^{n-1} \to X_{0}$ are homotopic equivalent, then the maps which they induce: $f, g : \bigsqcup_{\alpha} S_{\alpha}^{n-1} \to X_{0}$ are homotopic equivalent too. Hence, using the notations of the preceding theorem, we put $X_{1} := \bigsqcup_{\alpha} D_{\alpha}^{n}$, $A := \bigsqcup_{\alpha} S_{\alpha}^{n-1}$.

NOTE: by a relative CW complex $(X, A)$ I (and Spanier) mean simply a topological space $X$ obtained by attaching cells to $A$ in increasing dimension

This should be an exercise in spanier algebraic topology, but I can't find it anymore..


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