Equivalent definition of Cofibration

Question

Basic question from may's concise course about cofibration. In the beginning of chapter 6 (search pg 51 in this pdf: https://www.math.uchicago.edu/~may/CONCISE/ConciseRevised.pdf), May gives two equivalent diagrams, the second after applying the adjunction between cartesian product and mapping space. What I'm confused about is there's a map $p_0:Y^I \to Y$ defined by evaluation at time $0$, and i'm wondering why is the the correct map to make the diagram commute. why not evaluation at time $1, 1/2$, etc.? I'm sensing that we made a choice somewhere and I'm not seeing where

Answer 1

You know that there is 1-1-correspondence between homotopies $F : Z \times I \to Y$ and maps $e(F) : Z \to Y^I$ given by $e(F)(z)(t) = F(z,t)$. For the sake of clarity let us write in the second diagram $h'$ instead of $h$ and $\tilde{h}'$ instead of $\tilde{h}$.

Note that the lower solid arrow triangle in the first diagram does not admit possible commutativity relations and the outer square trivially commutes.

Let us check that the upper left triangle in the first diagram (obtained by omitting $X \times I$) commutes iff the outer square in the second diagram commutes when identifying $h' = e(h)$. We have $(p_0 e(h))(a) = p_0(e(h)(a)) = e(h)(a)(0) = h(a,0) = (h i_0)(a)$, thus $p_0 e(h) = h i_0$. This proves our claim but only works if we evaluate at $0$.

Let us next check that $\tilde{h}$ is a filler in the first diagram iff $e(\tilde{h})$ is a filler in the second diagram. As above we have $(p_0 e(\tilde{h}))(x) = \tilde{h}i_0(x)$, thus $p_0 e(\tilde{h}) = \tilde{h}i_0$. Hence $p_0 e(\tilde{h}) = f$ iff $\tilde{h}i_0 = f$. Moreover we have $\big((e(\tilde{h}) i)(a)\big)(t) = \big(e(\tilde{h})(i(a)\big)(t) = \tilde{h} (i(a),t) = \tilde{h} (i \times id)(a,t)$ and $\big(e(h)(a)\big)(t) = h(a,t)$. Hence $\tilde{h} (i \times id) = h$ iff $\big((e(\tilde{h}) i)(a)\big)(t) = \big(e(h)(a)\big)(t)$ which is equivalent to $e(\tilde{h}) i = e(h)$.