Is Every Cofibration Injective?

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SUMMARY

Every cofibration is injective, as established in the discussion surrounding the Problem of the Week (POTW). The concept of cofibrations is crucial in homotopy theory, particularly in the context of model categories. The discussion emphasizes the importance of understanding the properties of cofibrations to grasp their injective nature. No solutions were provided by other participants, highlighting the need for further exploration of this topic.

PREREQUISITES
  • Understanding of cofibrations in homotopy theory
  • Familiarity with model categories
  • Basic knowledge of injective objects in category theory
  • Experience with Problem of the Week (POTW) format
NEXT STEPS
  • Research the properties of cofibrations in homotopy theory
  • Study injective objects in category theory
  • Explore model categories and their applications
  • Review previous Problem of the Week discussions for context
USEFUL FOR

Mathematicians, particularly those focused on algebraic topology and category theory, as well as students seeking to deepen their understanding of cofibrations and injective objects.

Euge
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Here is this week's POTW:

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Why is every cofibration injective?

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Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!
 
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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.
 

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