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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 $S^1$.

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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

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