# Does the reduced Mapping cylinder have the same homotopy type of unreduced Mapping cylinder?

Let $f:X\to Y$ be a map in the pointed category of topological spaces $Top_*$. And let $U:Top_*\to Top$ be the "forgetful" functor (which "forgets" the basepoint). We can look at the reduced mapping cylinder $M_f$ and at the unreduced mapping cylinder $M_{U(f)}$ in the category $Top$. Until yesterday, I thought $M_f$ has the same homotopy type of $M_{U(f)}$. But, in May's Concise Course, he says "If $X, Y$ are well-pointed, then $M_f$ has the same homotopy type of $M_{U(f)}$. Is this hypothesis necessary?

If it is truly necessary, I want to know where I'm wrong. I thought this: Since $M_{U(f)}$ is a pushout of a trivial cofibration, we have that $Y\to M_{U(f)}$ is a trivial cofibration. The same way (or only using a explicit homotopy), we have that $Y\to M_f$ is a homotopy equivalence. So we have that $M_{U(f)}\equiv Y\equiv M_f$.

I know that the first statement is right. If there is something wrong, it is in the second statement. I believed that we can factor any function in $Top _*$ in the same way as in $Top$, id est, $f= R\circ j$, where $j: X\to M_f$ is a cofibration and $R: M_f\to Y$ is a strong retract. Is it wrong?

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Cross-posted on MO. – Alex Becker Sep 18 '12 at 2:47
@nunes: in general, please do not post a question on multiple sites simultaneously. – Qiaochu Yuan Sep 18 '12 at 3:44

Let $X$ be the disjoint union of $[0,1]$ and a single point $*$ and $Y=\mathbb{R}^1$. For any continuous map $f:(X,*)\rightarrow (Y,y_0)$, the unreduced mapping cylinder $M_{U(f)}$ is the same homotopy type of $Y$ while the reduced mapping cylinder $M_f$ is the same homotopy type of the wedge sum of $Y$ and the circle $S^1$.
Where are you getting that circle from...? In any case the inclusion $* \to X$ is a cofibration (it's a sub-CW-complex), so by general arguments the two should be the same...! – Najib Idrissi Sep 27 '14 at 14:01