Let $X$ and $Y$ be topological spaces and $f:X \to Y$ a continuous function. The mapping cylinder, $Cyl(f)$, is the push-out of: $$\begin{array} XX & \stackrel{f}\longrightarrow & Y \\\\ \downarrow{\iota_0} & & \\\\ X\times I & & \end{array}$$ Now, this gives the commutative diagram: \begin{array} AX & \stackrel{f}{\longrightarrow} & Y \\ \downarrow{\iota_0} & & \downarrow{j} \\ X \times I & \stackrel{\beta}{\longrightarrow} & Cyl(f) \end{array} Now, is $j$ an homeomorphism?

Notice that since the following diagram commutes ($p_X$ is the canonical projection over $X$): \begin{array} AX & \stackrel{f}{\longrightarrow} & Y \\ \downarrow{\iota_0} & & \downarrow{\text {id}_Y} \\ X \times I & \stackrel{f \circ p_X}{\longrightarrow} & Y \end{array} There's a unique map $r: Cyl(f) \to Y$ such that: $$r\circ j = {\text {id}_Y} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, r\circ\beta = f \circ p_X$$

Now, we may consider $j\circ r: Cyl(f) \to Cyl(f)$. This map satisfies: $$ (j\circ r)\circ j = j\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (j\circ r)\circ \beta = \beta$$ Since $Cyl(f)$ is a the push out of $(f,\iota_0)$, would the latter imply $j\circ r = {\text{id}}_{Cyl(f)}?$

I'm almost certain that there might be something wrong with my reasoning, but I am not seeing it right now. Any help would be appreciated.


Thanks to the suggestions and the example given by Mariano Suárez-Álvarez I realized my mistake. When calculating $(j \circ r)\circ \beta$, I did: \begin{align*} (j \circ r)\circ \beta &= j\circ (r\circ \beta) \\ &= j\circ ( f \circ p_X) \\ &= (j \circ f) \circ p_X \\ &= (\beta \circ \iota_0) \circ p_X \end{align*} I assumed $\iota_0 \circ p_X = {\text {id}_{X\times I}},$ and this is false. It goes the other way: $p_X \circ \iota_0 = {\text{ id}_{X}}.$

Other Edit:

In fact; since $I$ is contractible, $\iota_0 \circ p_X$ is homotopic to ${\text {id}_{X\times I}}$. Then $p_X$ and $\iota_0$ are an homotopy equivalence. By the reasoning above, we get that $r$ is a homotopy equivalence.

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    $\begingroup$ Have you tried to construct any example of a mapping cylinder to see where your proof fails? $\endgroup$ – Mariano Suárez-Álvarez Oct 21 '16 at 4:21
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    $\begingroup$ By the way, make yourself a favor and read the documentation of amsmath! $\endgroup$ – Mariano Suárez-Álvarez Oct 21 '16 at 4:22
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    $\begingroup$ Consider the constant map from [0,1] to a one-point space: what is the mapping cylinder? $\endgroup$ – Mariano Suárez-Álvarez Oct 21 '16 at 4:27
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    $\begingroup$ I don't know exactly what you mean by «I am using only categorical properties » but I suggest that if that is in any way an obstacle in your finding examples of things, then stop using only that! :-) The people that came up with all these definitions did not pull them out of thin air, but out of very concrete examples in topological spaces and so in, and then used the categorical language to streamline that... $\endgroup$ – Mariano Suárez-Álvarez Oct 21 '16 at 4:29
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    $\begingroup$ Indeed, a triangle, and it is not isomorphic to anything in sight. Now go through your argument and see what breaks. $\endgroup$ – Mariano Suárez-Álvarez Oct 21 '16 at 4:46

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