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This is the problem:

Let $X=\{a,b\}$ be the topological space with trivial topology and $A=\{a\}$ the one point set (obviously not open). Prove using definition of cofibration that inclusion $A\to X$ is cofibration.

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    I'm trying to construct extended homotopy F' but I get stuck because I get that the extended homotopy F' is equal to F on A, and on X\A it can be any continuos function2012-10-09
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    Well, yes, and what's wrong with it? The simplest if you define the same for $b$ as what is given for $a$.2012-10-09
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    Isn't that too easy? Then I do nothing special, just define F'(x,t)=F(a,t) for every x in X and t in [0,1]. Shouldn't I prove something?2012-10-09
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    That it is *continuous*, and that the diagrams really commute.2012-10-09
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    But it's obvious, or not? F is homotopy, so it's continuous and then F' is also continuos, and another when I use that definiton for F' diagram really commute.2012-10-09
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    then you're done. I think, $b$ must be mapped the same as $a$ for continuity, but that's all. Go on to next exercise.2012-10-09

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If $H:A\times I\to Y$ is a homotopy and $f:X\to Y$ is a map such that $H_0|_A=f$, you could define $$H':X\times I\to Y\\ H'(x,t)=\begin{cases}H(a,t), &\text{if }t>0\\ f(x), &\text{if }t=0 \end{cases}$$ On $X\times(0,1]$ this is continuous, since it is just the composition $(x,t)\mapsto(a,t)\mapsto H(a,t)$.

$H'$ is continuous at $(a,0)$: Since $f$ is continuous and $a\in\overline{\{b\}}$, $\ f(b)=H'(b,0)$ is element of each neighborhood $V$ around $f(a)$. Since $H$ is continuous, there is an $\varepsilon>0$ such that $H(A\times[0,\varepsilon))$ is mapped to $V$. But then also $X\times[0,\varepsilon)$ is mapped to $V$ by $H'$.

$H'$ is continuous at $(b,0)$: By the argument above, $f(a)\in V$ for each open neighborhood $V$ around $f(b)$, and for some $\varepsilon>0$, the set $A\times[0,\varepsilon)$ is mapped to $V$. Therefore $H'(X\times[0,\varepsilon))\subseteq V$.