I have constructed the following proof of the statement and have some questions (a question) about the correctness of the proof:


$A$ retract of $X$ and $X$ contractible implies $A$ contractible.


I make the following observations:

  • Since $X$ is contractible we have $1_{X} \simeq constant $. This follows from a proposition that a space $X$ is contractible if and only if $1_{X}$ is nullhomotopic.
  • $A$ retract of $X$ gives us that there exists a map $r: X \rightarrow A$ such that $r\circ i = 1_{A}$ where $i:A\rightarrow X$ is the inclusion map.

Now I conclude:

$1_{A} = r\circ i = r \circ 1_{X} \circ i \simeq r \circ constant \circ i$

Since $r \circ constant \circ i$ is a constant map we have that $1_{A}$ is nullhomotopic and thus we have $A$ is contractible.


  • Is this proof correct?
  • I use that $f \simeq f_{1}$ implies $g \circ f \simeq g\circ f_{1}$. Is this true? I think it makes sense.

When you write $r\circ 1_X\circ i\simeq r\circ \text{constant}\circ i,$ then you implicitely mean that the homotopy is $r\circ F\circ(i\times 1_I),$ so your proof is correct.

In general, if $f_1\simeq f_2$ by $F$ then $g\circ f_1\simeq g\circ f_2$ by $g\circ F,$ and $f_1\circ h\simeq f_2\circ h$ by $F\circ(h\times 1_I).$


Here is an alternative proof, demanding some familiarity with categories.

The contractible objects in category $\mathbf{Top}$ are exactly the terminal objects in category $\mathbf{hTop}$. If $X$ is a terminal object in $\mathbf{hTop}$, then for every object $Y$ the homset $\mathbf{hTop}\left(Y,X\right)$ contains exactly one arrow. Let us denote this arrow by $\left[c\right]$. Then homset $\mathbf{hTop}\left(Y,A\right)$ contains arrow $\left[r\right]\left[c\right]$. Now let $\left[f\right]\in\mathbf{hTop}\left(Y,A\right)$. Then $\left[i\right]\left[f\right]\in\mathbf{hTop}\left(Y,X\right)$ so $\left[i\right]\left[f\right]=\left[c\right]$. Then we find $\left[f\right]=\left[r\right]\left[i\right]\left[f\right]=\left[r\right]\left[c\right]$ demonstrating that $\left[r\right]\left[c\right]$ is the only element of $\mathbf{hTop}\left(Y,A\right)$. Proved is now that $A$ is terminal in $\mathbf{hTop}$, hence contractible in $\mathbf{Top}$.

In my view this proof is a good demonstration of 'the joy of cats'.

Later: A bit shorter:

$\left[\right]:\mathbf{Top}\rightarrow\mathbf{hTop}$ is a functor and functors respect retractions (=arrows that have a right-inverse). So if $r:X\rightarrow A$ is a retraction in $\mathbf{Top}$ then $\left[r\right]:X\rightarrow A$ is a retraction in $\mathbf{hTop}$.

If $\left[r\right]:X\rightarrow A$ is a retraction and $X$ is terminal then $\left[r\right]$ is an isomorphism, hence $A$ is terminal. (This is true in every category, but to maintain the line of the proof I keep on using the notation $\left[r\right]$.)

(Proof: some $\left[s\right]:A\rightarrow X$ exists with $\left[r\right]\left[s\right]=1$; then $\left[s\right]\left[r\right]:X\rightarrow X$ and the identity is the only endomorphism here, so $\left[s\right]\left[r\right]=1$.)

  • $\begingroup$ Leuk! (=nice in Dutch). Can you suggest some source where homotopy theory or $hTop$ or algebraic topology are developed using the language of categories? $\endgroup$ – magma Sep 26 '13 at 14:30
  • $\begingroup$ @magma Ja, leuk hè. Lekker kort! Sorry, I cannot help you with any source. I just found out myself. $\endgroup$ – drhab Sep 26 '13 at 14:33
  • $\begingroup$ Jammer! (what a pity!) Check my profile if you like to en-joy the cats with a Computer Algebra System (Mathematica). $\endgroup$ – magma Sep 26 '13 at 14:48
  • $\begingroup$ @magma I allready did that. However I have no Mathematica on my computer, wich is required for this. $\endgroup$ – drhab Sep 26 '13 at 14:51
  • $\begingroup$ you may contact me per email if you would like to discuss this further $\endgroup$ – magma Sep 26 '13 at 15:10

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