Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The pushout in the category of topological spaces is not a homotopy invariant. Can somebody give me an explicit example for this fact?

share|improve this question
add comment

1 Answer 1

up vote 7 down vote accepted

The pushout of $I \leftarrow 2 \rightarrow I$ where $I = [0, 1], 2 = \{ 0, 1 \}$, and both maps are the inclusion of the endpoints into the interval is $S^1$. This pushout diagram is homotopy equivalent to $1 \leftarrow 2 \rightarrow 1$ (where $1$ is a one-point space) whose pushout is $2$, which is not homotopy equivalent to $S^1$.

share|improve this answer
    
Thank you very much! An additional question: Is the pullback also not a homotopy invariant? –  Your Ad Here Oct 17 '12 at 18:10
    
There's no particular reason for it to be. –  Qiaochu Yuan Oct 17 '12 at 18:12
4  
Indeed, a very similar example works: there's a cospan $1 \rightarrow I \leftarrow 1$ whose pullback is empty, but contracting $I$ to a point first makes the pullback $1$ instead. –  Zhen Lin Oct 17 '12 at 18:35
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.