Given a continuous map $f\colon X\to Y$ between two non-empty topological spaces, show that there is homotopy equivalence between the mapping cylinder $(X\times I)\sqcup _{f}Y$ and Y.

Here we have I=$[0,1]$, $ (x,1)\sim f(x)$ on X.

Is the following proof correct?:

Let's denote by $\cong$ a homotopy equivalence.

I is contractible so $I\cong \{1\}$, and obviously $X \cong X$,

So we have $X\times I \cong X\times \{1\} \cong f(X)$

Therefore $X\times I \sqcup_f Y\cong f(X)\sqcup_f Y = Y$

Thank you for your corrections and comments.

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    $\begingroup$ If $X$ is noncontractible and $f$ is constant then $X \times \{1\} \cong f(X)$ is false. So no, the proof is not correct. $\endgroup$ – Lee Mosher Feb 17 at 17:02
  • $\begingroup$ Right! Thank you for the counter-example. Is homotopy equivalence compatible with the product of spaces as I used it for this statement $X\times I \cong X\times \{1\}$? I know it is the case for deformation retraction $\endgroup$ – PerelMan Feb 17 at 17:08
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    $\begingroup$ Note that you do not have $(x,0)\sim(x',0)$ for the mapping cylinder. If you had that relation, you'd get the mapping cone, which does not retract onto $Y$ in general. $\endgroup$ – Christoph Feb 17 at 17:36
  • $\begingroup$ Thanks! I edited the question $\endgroup$ – PerelMan Feb 17 at 17:46

Denote $M :=(X\times I)\sqcup _{f}Y$ and take maps $p: Y \to M, y \mapsto y$ and $q: M \to Y, (x, s) \mapsto f(x), y \mapsto y$. Well defined: easy to verify

Then $q \circ p= id_Y$ and $p \circ q \cong id_M$ via the homotopy map

$$H_t: M \times I \to M, (x, s,t) \mapsto (x, t +s(1-t)), y \mapsto y$$

Then: $H_0= id_M, H_1= p \circ q$

  • $\begingroup$ if $t\in [0,1[$ then $q(x,t) \not \in Y$ so $q$ is not defined all over M? $\endgroup$ – PerelMan Feb 17 at 17:29
  • $\begingroup$ sure, $q$ maps by definition $(x,t)$ to $f(x) \in Y$. take futhermore into account that $H_1$ factorize by construction through $Y$, since $X \times \{1\} \cong f(X) \subset Y$ by your identifications $\endgroup$ – KarlPeter Feb 17 at 17:32
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    $\begingroup$ one subtle remark: $X \times \{1\} \cong f(X)$ holds only in $M$, so only if we consider the both as subsets of $M$. More precisely: $X \times \{1\}/\sim_M \cong f(X)/\sim_M$. here $\sim_M$ are the identifications in $M$ $\endgroup$ – KarlPeter Feb 17 at 17:42
  • $\begingroup$ Many thanks Sir ;) $\endgroup$ – PerelMan Feb 17 at 17:55

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