MHB Is Every Cofibration Injective?

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Every cofibration is injective due to its properties in homotopy theory, which ensure that cofibrations allow for the extension of maps. The discussion highlights the significance of cofibrations in topological spaces and their role in preserving homotopy types. Despite the complexity of the topic, no participants provided answers to the posed problem. A solution was shared by the original poster, emphasizing the theoretical underpinnings of cofibrations. The thread illustrates the challenge of engaging with advanced mathematical concepts in community discussions.
Euge
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Here is this week's POTW:

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Why is every cofibration injective?

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No one answered this week's problem. You can read my solution below.
Let $f : X' \to X$ be a cofibration, and let $M_f$ denote the mapping cylinder of $f$. Consider the projection map $g : X \to Z_f$ and the map $G : X' \times I \to M_f$ sending $(x',t)$ to the equivalence class of $(x',1-t)$. For every $x'\in X'$, $gf(x') = [f(x')] = [(x',1)] = G(x',0)$. Since $f$ is a cofibration, there is a map $\Phi : X\times I \to M_f$ such that $\Phi(x,0) = [x]$ for all $x\in X$ and $\Phi(f(x'),t) = [(x',1-t)]$ for all $x'\in X$ and $t\in I$. Hence, given $a',b'\in X'$ with $f(a') = f(b)'$, we have $\Phi(f(a'),1) = \Phi(f(b'),1)$, or $[(a',0)] = [(b',0)]$. Therefore $a' = b'$, showing that $f$ is injective.