The claim that $A \to X$ a cofibration implies $A \times I \to M_{A \to X}$ is an inclusion.

228 Views Asked by At

Akhil Matthew claims in https://amathew.wordpress.com/2010/10/07/cofibrations/ in parenthesis that given a cofibration, $A \xrightarrow{i} X$, the map $A \times I \to M_i$ into the mapping cylinder, is an inclusion.

The claim that the map $A \times I$ into the mapping cylinder $M_i=A \times I \sqcup_f X$ is an inclusion whenever $i: A \to X$ is a cofibration seems dubious.

This map is an inclusion whenever the map $A \to A \times 1 \to A \times I \sqcup _f X$ is an inclusion, which happens whenever $i$ is injective.

Thus for any noninjective map $A \xrightarrow{j} X$, we have the map $\phi: A \times I \to A \to A \sqcup X_f$. Then $A \cong A \times 0 \to A \times I \xrightarrow{\phi} \to A \sqcup X /(a ~ f(a)) \hookrightarrow M_i$ is a cofibration such that $A \times I \to M_i$ is not an inclusion.

Is my assesment correct?