# Is this inclusion a cofibration?

Let $$(X,A)$$ be a relative CW-complex. Consider the inclusion $$i:\ \ X\times \{0,1\} \cup A\times I \rightarrow X\times I.$$

I was wondering if this is a cofibration.

I guess it is, for there is a push out diagram $$\require{AMScd}$$ $$\begin{CD} A\times (0,1) @>>> X\times \{0,1\} \cup A\times I\\ @VVV @VVV\\ X\times(0,1) @>>> X\times I \end{CD}$$ the left vertical arrow is obviously a cofibration. So is the right vertical arrow.

Is the above argument correct?

• Why is this diagram a pushout diagram? – Paul Frost Feb 13 at 18:40
• @PaulFrost Since $X\times I$ is the disjoint union of $X\times (0,1)$ and $X\times \{0,1\}\cup A\times I$ after identifying $A\times (0,1)$. Since all the maps are inclusions, the diagram commutes. – Aolong Li Feb 13 at 18:46

Your argument is not correct because the diagram in your question is not a pushout diagram. If it were one, then for any two maps $$\phi : X \times \{ 0, 1 \} \cup A \times I \to Y$$ and $$\psi : X \times (0,1) \to Y$$ which agree on $$A \times (0,1)$$ we would get a unique map $$\chi : X \times I \to Y$$ whose restriction to $$X \times \{ 0, 1 \} \cup A \times I$$ is $$\phi$$ and whose restriction to $$X \times (0,1)$$ is $$\psi$$.

Counterexample 1.

Any CW-complex $$X$$ can be regarded as relative CW-complex $$(X,A) = (X,\emptyset)$$. Take $$Y = I$$ and define $$\phi : X \times \{ 0, 1\} \to I, \phi(x,t) = 0$$ and $$\psi : X \times (0,1) \to I, \psi(x,t) = t$$. These maps agree on $$\emptyset \times (0,1)$$ but do not give you the desired $$\chi : X \times I \to I$$.

Counterexample 2.

Let $$(X,A) = (I,\{ 0\})$$. Define $$\phi : I \times \{ 0, 1 \} \cup \{ 0 \} \times I \to I, \phi(x,y) = y$$ and $$\psi : I \times (0,1) \to I, \psi(x,y) = \min (x+y,1)$$. These maps agree on $$\{ 0 \} \times I$$ but again you do not the desired $$\chi : I \times I \to I$$.

A correct proof can be based on the observation that $$(X \times I, X \times \{ 0, 1 \} \cup A \times I)$$ is also a relative CW-complex.

By the way, it is a general theorem that if an inclusion $$A \hookrightarrow X$$ is a cofibration, then so is $$X \times \{ 0, 1 \} \cup A \times I \hookrightarrow X \times I$$. The proof requires some deeper properties of cofibrations. See Theorem 6 of

Strøm, Arne. "Note on cofibrations II." Mathematica Scandinavica 22.1 (1969): 130-142.