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Let $Y$ be a normal space, we say that it is ANE (Absolute Neighborhood Extensor) if for every metric space $X$ and closed subset $A$ of $X$, if $h\colon A\to Y$ then there is an open neighborhood $U$ of $A$ that we can extend $h$ to $h'\colon U\to Y$.

We say that $Y$ has HEP (Homotopy Extension Property) if for every metric space $X$ and closed subset $A$ of $X$, if $h\colon(X\times\{0\})\cup(A\times[0,1])\to Y$ is continuous then we can extend it to $H\colon X\times[0,1]\to Y$.


I want to show that $Y$ is ANE then $Y$ has HEP. So let $Y$ be ANE and $X$ a metric space, $A\subseteq X$ closed.

Suppose $F=(X\times\{0\})\cup(A\times[0,1])$ and $h\colon F\to Y$ is continuous. We can therefore find an open $U\supseteq F$ and extend $h$ to $h'\colon U\to Y$. By Urysohn's lemma there is $g\colon X\times[0,1]\to[0,1]$ continuous such that: $g(x,t)=1$ for $(x,t)\in F$ and $g(x,t)=0$ for $(x,t)\notin U$.

Define $H(x,t)=h'\Big(x,\ g(x,t)\cdot t\Big)$. Clearly $H$ extends $h$ continuously, however there is a fine detail which I cannot overcome:

Suppose $(x,t)\in U$ why does $(x,\ g(x,t)\cdot t)\in U$ as well? I cannot think of a reason why this should be true, but I also cannot find an argument why this must not be possible to have (in a sense, $U$ should be closed downwards in the second coordinate).

Any hints, tips or corrections to my suggested solution above are welcomed.

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    Will the gracious downvoter care to explain?2012-02-06
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    What if you first replace this $U$ with some open $U'\subseteq U$ that still contains $A$ and is "downward closed in the second component"? Restrictions of continuous maps remain continuous, so the restriction $h'|U'$ will remain continuous.2012-02-06
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    A thought which may or may not be useful: if $A$ is compact, you can choose $U$ to be a basic neighbourhood, so there isn't a problem in this case.2012-02-06
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    @Dejan: Of course, I just can't come up with a good excuse for replacing $U$ with this $U'$. I just don't see why it *has* to exist.2012-02-06
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    @Miha: $A$ itself is not necessarily compact.2012-02-06
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    Yes, I was just musing on one possible case. Regardless, I think the following should work: for each $t\in[0,1]$ the distance from $A_t$ to the boundary of $U_t$ is positive, where $A_t,U_t$ are "slices" at time $t$. By compactness, there is a minimal positive distance. Using this minimum, construct a basic neighbourhood of $F$, contained in $U$.2012-02-06
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    @Miha: I thought about that but couldn't find the exact argument (I neglected the compactness of $[0,1]$), if you want to puff up the details and write an answer I'll accept.2012-02-06
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    Unfortunately, this doesn't seem to pan out. I'll think about it some more, but I have a creeping feeling that weird things can happen with boundaries of open sets.2012-02-06

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I believe I have found an argument that works.

Let $U$ be an open set such that $(X\times\lbrace 0\rbrace)\cup (A\times[0,1])\subseteq U$. Define $\mathcal{U}=\lbrace V\times W|\hbox{ }V \textrm{ is open in }X,\hbox{ } W \textrm{ is open in }[0,1] \textrm{ and }V\times W\subseteq U\rbrace$, the set of all boxes contained in $U$.

Lemma. For every $a\in A$ there exists a downward closed open neighborhood $U_a$ of $\lbrace a\rbrace\times[0,1]$ such that $U_a\subseteq U$.

(Here of course, a set $U'\subseteq X\times[0,1]$ is defined to be downward closed if for every $(x,t_1)\in U'$ and every $t_2\in[0,t_1]$ we have $(x,t_2)\in U'$.)

Proof. Let $a\in A$. Then $\lbrace a\rbrace\times[0,1]$ is compact, so there exist $V_1\times W_1,V_2\times W_2,...,V_n\times W_n$ in $\mathcal{U}$ such that $\lbrace a\rbrace\times[0,1]\subseteq\bigcup_{i=1}^n V_i\times W_i$. But $V = \bigcap_{i=1}^n V_i$ is a finite intersection of open sets that all contain $a$ (or at least we can safely remove those that do not contain $a$ and still get an open cover), so it must itself be an open set that contains $a$. Thus we see that $V\times\bigcup_{i=1}^n W_i = V\times[0,1]$ is an element of $\mathcal{U}$ and downward closed which proves the lemma.

Next, define $\mathcal{L}=\lbrace V\times W\in \mathcal{U}|\hbox{ }\exists t\in[0,1]:W=[0,t)\rbrace$ and $U_\infty=\bigcup\mathcal{L}$.

Observe that an arbitrary union of downward closed sets is again downward closed. Therefore $U_\infty$ is a downward closed set that contains $X\times\lbrace0\rbrace$.

Now just take $U'' = U_\infty\cup\bigcup_{a\in A} U_a$. This is a union of downward closed sets and therefore downward closed. It is contained in $U$ because every set in the union is contained in $U$. It contains $X\times\lbrace 0\rbrace$ and $\lbrace a \rbrace\times[0,1]$ for every $a\in A$. So it also contains $(X\times\lbrace 0\rbrace)\cup (A\times[0,1])$ and the problem is solved.

Any corrections/comments are welcome.

Edit: for better readability I shall state the conclusions here. The above argument hopefully shows that every open neighborhood of $F = (X\times\lbrace 0\rbrace)\cup (A\times[0,1])$ contains a downward closed open neighborhood of $F$. Since the extension $h':U\to Y$ of $h: F\to Y$ is continuous, so is its restriction to this smaller neighborhood $h'|_{U''}:U''\to Y$. Therefore this is also an extension of $h$. Since now $U''$ is downward closed, $H$ will now be well defined.

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    Why is $V\times\bigcup_{i=1}^n W_i$ downwards closed?2012-02-07
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    Because $\bigcup_{i=1}^n W_i = [0,1]$. (Since $(V_i\times W_i)_{i=1}^n$ is a cover of $\lbrace a\rbrace\times[0,1]$.)2012-02-07
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    So we can just write that instead :-) $V\times[0,1]$ is clearer! Thanks.2012-02-07
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    Yeah, you are right. It was a bit late when I was typing that =) I have added that. I hope it is clearer now. =)2012-02-07