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

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1 Answer 1

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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 $S^1$.

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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
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

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