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I have to show that if $X$ is a CW-complex, $A$ is a subcomplex and $i:A\hookrightarrow X$ is a cellular inclusion, then $i$ is a cofibration. My attempt is as follows. I think my proof, which first proves a lemma and is based on an induction argument, looks kind of cluttered. I relaize the proof can be made shorter by using theorems about NDR-pairs, but this problem appears before they are introduced, so I don't think they are "allowed". I have two questions:

  1. Is my proof correct? CW-complexes are pretty well-behaved, so I doubt there can be many pitfalls, but you never know...

  2. Is there a better way of proving this, without using material about NDR-pairs.

My strategy is to show that $j:S^{n-1}\rightarrow D^{n}$ is a cofibration for all $n\geq 1$, and prove that a homotopy in $A$ extends inductively on each $X^k$.

So assume we have a homotopy $h^\prime:S^{n-1}\times I\rightarrow Y$ and $\bar{f}:D^{n}\rightarrow Y$ such that $h^\prime(x,0)=\bar{f}\circ j(x)$ for all $x\in S^{n-1}$. Then define $\bar{h}:D^{n}\times I\rightarrow Y$ by $$\bar{h}(x,t)=\left\{ \begin{array}{lcl} \bar{f}\left( (t+1)x\right) & , & ||x||\leq \frac{1}{t+1} \\ h^\prime\left(\frac{x}{||x||},(t+1)||x||-1\right) & , & ||x|| \geq \frac{1}{t+1} \end{array} \right.$$ Because $\bar{h}\circ j(x,t)=h^\prime(x,t)$, this extends $h$ to a homotopy $\tilde{h}:D^{n}\rightarrow Y$.

So now we have shown that $j:S^{n-1}\rightarrow D^n$ is a cofibration. Now let $h:A\times I\rightarrow Y$ be a homotopy and $f:X\rightarrow Y$ such that $h(x,0)=f\circ i(x)$ for all $x\in A$. We construct $\tilde{h}:X\times I\rightarrow Y$ inductively on the skeletons $X^k$ of $X$.

Base case: $X^0$ is trivial: define $\tilde{h}=\left\{ \begin{array}{lcl} h(x,t) & , & x\in A \\ f(x) & , & x\not\in A \end{array} \right.$.

Assume $\tilde{h}$ is constructed on $X^{k-1}$. For each $k$-cell $E^k$ in $X$, if $E^k\subseteq A$, then define $\tilde{h}(x,t)=h(x,t)$ for all $(x,t)\in A\times I$. If $E^k\not\in A$, then let $g:D^k\rightarrow X$ be the attaching map of $E^k$. Define $f\circ g=\bar{f}:D^k\rightarrow Y$ and $h^\prime:S^{k-1}\times I$ such that $h^\prime=h\circ (g\times \mathrm{id})$. Then we have the same situation as above, and $h^\prime$ extends to a homotopy $\bar{h}:D^k\times I \rightarrow Y$. Then $\bar{h}$ restricted to $\mathrm{int}\,\,D^k$ extends the homotopy $h$ to $A\cup X^{k-1} \cup E^k$. When we do this for every $k$-cell, since the extended homotopies agree on $X^{k-1}$, we get a well-defined extension $\tilde{h}:A\cup X^k\rightarrow Y$.

This completes the inductive step and we have proved that $h:A\times I\rightarrow Y$ can be extended to $\tilde{h}:X\times I \rightarrow Y$, and consequently that $i$ is a cofibration.

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