$i$ is a cofibration then $1 \times i : B \times A \rightarrow B \times X$ is cofibration

algebraic-topologyhomotopy-theoryproof-explanation

Let $i:A \rightarrow X$ be a cofibration and let $B$ be a space. If $i$ is closed, or $B$ is locally compact, then $1 \times i: B \times A \rightarrow B \times X$ is a cofibration.

I have found a proof online, but I am struggling in understanding Ronald Brown, pg267, 7.2.4(Corollary 2) .

What I understand:

  1. It is equivalet to prove, by 7.2.4 pg266, that $\mu':M(1 \times i) \rightarrow B \times X \times \mathbb{I}$ is a coretraction.

My questions:

  1. Why does proving the natural map $M(1 \times i) \rightarrow B \times M(i)$ is a homeomorphism implies it is a coretraction?

  2. In the case when $i$ is a closed fibration, why does the pushout diagram implies the first diagram is a push out?

Best Answer

  1. A co-retraction is defined as a map having a left inverse. We know that $\mu : M(i) \to X \times \mathbb{I}$ is co-retraction with retract $\rho$. Therefore $1_B \times \mu : B \times M(i) \to B \times X \times \mathbb{I}$ is a co-retraction with retract $1_B \times \rho$. This shows that also $\mu' : M(1_B \times i) \to B \times X \times \mathbb{I}$ is a co-retraction.

  2. If you accept that the second diagram on p. 268 is a pushout diagram, then also the first diagram ($\ast$) is one simply because you have homeomorphisms between the four spaces at the corners of the two squares producing a commutative cubical diagram.

By the way, $1_B \times i$ is always a cofibration if $i$ is one. The general proof is much harder than the proof based on the assumptions in your question.

Added:

The natural map $h : M(1_B \times i) \to B \times M(i)$ is constructed via the universal property of the pushout.

By definition, $M(1_B \times i)$ occurs in lower right corner of a pushout diagram - that we denote for the moment by ($\ast\ast$) - based on the two maps $B \times A \times 0 \to B \times X \times 0$ and $B \times A \times 0 \to B \times A \times \mathbb{I}$ of diagram ($\ast$).

Similarly $M(i)$ occurs in a pushout diagram based on the two maps $A \times 0 \to X \times 0$ and $A \times 0 \to A \times \mathbb{I}$. Multiplying this diagram from the left (Cartesian product!) with $B$ produces the commutative diagram ($\ast)$. The universal property of the pushout gives you a unique map $h : M(1_B \times i) \to B \times M(I)$ with the well-known properties.

Ronnie Brown proves that ($\ast)$ is pushout diagram. Since both ($\ast\ast$) and ($\ast$) are pushout diagrams, the map $h$ must be a homeomorphism.

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