Proof that $X \subset X \times I$ is a cofibration.

Question

I'm trying to see intuitively that $X \subset X \times I$ is a cofibration.

One approach i can think of is to use the result stated for example in this question The product of a cofibration with an identity map is a cofibration and apply it to $\emptyset \subset I$ which is trivially a cofibration.

That being said this seems to be a very simple example and i would like to be able to either see this straight from the definition of cofibration or by finding the appropriate retract. Is there a simpler way than i proposed? Is that method valid?

score 1 · Accepted Answer · answered Aug 16 '17 at 22:50

I think you meant $* \to I$ instead of $\emptyset \to I$ in your argument, but the rest of it looks fine.

Anyway, I don't think finding a retract (which also lets you write down an explicit extension in the definition of a cofibration) is "simpler" than your method, but here's how.

We need to show that $$(X \times I \times \{0\}) \cup_{X \times \{0\} \times \{0\}} (X \times \{0\} \times I) \subset X \times I \times I$$ is a retract. We can build such a retract from $(I \times \{0\}) \cup_{(0,0)} (\{0\} \times I) \subset I \times I$. Consider the homotopy $$H_t(u,v) = \begin{cases} (t(u-v) + (1-t)u, (1-t)v) & u \geq v \\ ((1-t)u, t(v-u) + (1-t)v) & u \leq v. \end{cases}$$ This is a deformation retraction of $I \times I$ onto $(I \times \{0\}) \cup_{(0,0)} (\{0\} \times I)$. This is essentially a proof that $\{0\} \to I$ is a cofibration.

Then $K_t(x,u,v) = (x, H_t(u,v))$ defines a deformation retract of $X \times I \times I$ onto $(X \times I \times \{0\}) \cup_{X \times \{0\} \times \{0\}} (X \times \{0\} \times I)$.

Great answer, just one followup question: by * in the first line you mean a single-element space? I'm not familiar with the notation. — TheMountainThatCodes, Aug 17 '17 at 19:51
Yes. The product of ${0} \to I$ with $X$ is $X \cong X \times {0} \to X \times I$. — JHF, Aug 17 '17 at 20:01

score 1 · Answer 2 · answered Aug 17 '17 at 12:41

A more general result is Strom's theorem: An inclusion $A\hookrightarrow Y$ is a cofibration if and only if there is a function $\phi:Y\rightarrow I$ such that $A=\phi^{-1}(0)$ is a strong neighbourhood deformation retract of $\phi^{-1}([0,1))$.

Applied to your case, take $A=X\times 0$, $Y=X\times I$ and $\phi=pr_2$, and the rest is clear.

It may not be intuitive, but the proof of Stroms result, which you may like to look up, certainly provides intuition and a deeper understanding.

Proof that $X \subset X \times I$ is a cofibration.

2 Answers2