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\noindent{\bf Prob. 3, p. 213.}
Let $X$ be a space and let $X_C$ denote $X$ with the topology
coherent with the collection of compact subsets..
\item{(a) }If $B\subset X$, show $B$ is compact as a subspace of
$X$ if and only if it is compact as a subspace fo $X_C$.

\pf
Let $B$ denote the set $B$ with the subspace topology from $X$
and use $B_C$ to denote $B$ with the subspace topology from $X_C$.
The function $i\colon X_C\to X$ which is the identity as a map of sets
is continuous.
To see this, let $A\subset X$ be closed.
We need to see $i^{-1}(A)\subset X_C$ is closed, which is equivalent
to checking that $A\cap C$ is a closed subset of $C$ for every compact
subset $C\subset X$.
But $A\cap C$ is displayed as $C$ intersected with a closed subset of $X$
so it is closed in the subspace topology on $C$.
Now $i$ restricts to a continuous bijection $B_C\to B$.
If $B_C$ is compact, so is $B$ since it is the image of a compact set.
Now suppose $B$ is compact.
If $A\subset B_C$ is closed, this means there exits $\bar A\subset X_C$
which is closed so that $A=\bar A\cap B_C$.
But $\bar A$ is closed in $X_C$ so $\bar A\cap D$ is closed
in $D$ for every compact subset $D\subset X$.
Taking $D=B$, we see
$\bar A\cap B$ is closed in $B$.
But $\bar A\cap B=A$, so $A$ is closed in $B$.
Hence $i\colon B_C\to B$ is a closed map, hence a homeomorphism.
Since $B$ is compact, so is $B_C$.
\medskip
A  space is called {\sl compactly generated\/} provided the
natural map $i_X\colon X_C\to X$ is a homeomorphism.

\item{(b)}$X_C$ is compactly generated.

\pf
We show the continuous bijection 
$i_{X_C}\colon (X_C)_C\to X_C$ is closed which shows that
it is a homeomorphism.
Let $A\subset (X_C)_C$ be closed.
This means $A\cap D$ is closed in $D$ for every compact
subset $D\subset X_C$.
By part(a), every compact subset $E\subset X$ is compact in
$X_C$ so $A\cap E$ is closed in $E$.
By definition, $A$ is closed in $X_C$.

\medskip\noindent{\bf Remark:}
Let $f\colon Y\to X$ is a map.
Let $f_C\colon Y_C\to X_C$ be the map induced
on the sets by $f$.
Then $f_C$ is continuous.
To show this, let $A\subset X_C$ be closed and consider
$f_C^{-1}(A)=f^{-1}(A)$.
Let $D\subset Y$ be compact: then so is $f(D)$.
Observe $f^{-1}(A)\cap D= f^{-1}\bigl(A\cap f(D)\bigr)\cap D$.
Now $A\cap f(D)$ is closed in $f(D)$ so $ f^{-1}\bigl(A\cap f(D)\bigr)$
is closed in $f^{-1}\bigl(f(D)\bigr)$.
It follows that $f^{-1}\bigl(A\cap f(D)\bigr)\cap D$ is closed in
$f^{-1}\bigl(f(D)\bigr)\cap D$, hence $f^{-1}(A)\cap D$ is closed in $D$.
But this means $f^{-1}(A)$ is closed in $Y_C$ and hence $f_C$ is
continuous.
\medskip
\item{(c)}Show that the inclusion $i_X\colon X_C\to X$ induces
an isomorphism in singular homology.

\pf
Consider the chain map $(i_X)_\#\colon S_\ast(X_C)\to S_\ast(X)$.
Since $i_X$ is one to one, so is $(i_X)_\#$.
To see that $(i_X)_\#$ is onto, let $T\colon \Delta^r\to X$
be any singular simplex.
By part (a), $\Delta^r_C=\Delta^r$ and the composition
$\vbox{\vskip 12pt}\Delta^r\ \RA{i^{-1}_{\Delta^r}}\ \Delta^r_C\ 
\RA{T_C}\ X_C\ \RA{i_X}\ X$ 
is just $T$.
Hence $(i_C)_{\#}(T_C)=T$ so $(i_C)_{\#}$ is onto.
Since the chain complexes are isomorphic via $(i_C)_\#$,
$(i_X)_\ast$ induces an isomorphism in singular homology.
\medskip
\item{(d)}If $X\times_CY=(X\times Y)_C$, and if $K_1$ and $K_2$ are
simplicial complexes, show that $\vert K\vert\times_C\vert L\vert$
is coherent with the subspaces $\sigma\times \tau$.

\pf
By Lemma 2.5 p.10 of the book, any compact subset of $\vert K\vert$
or $\vert L\vert$ is contained in a finite subcomplex.
If $D\subset \vert K\vert\times\vert L\vert$ and
if $\pi_K\colon \vert K\vert\times\vert L\vert\to\vert K\vert$ and
$\pi_L\colon \vert K\vert\times\vert L\vert\to\vert L\vert$ denote
the projections, $D\subset \pi_K(D)\times \pi_L(D)$.
Hence if $D$ is compact so are $\pi_K(D)$ and $\pi_L(D)$ and hence
there are finite subcomplexes $K_0$ and $L_0$ so that
$D\subset\vert K_0\vert\times\vert L_0\vert$.
Moreover 
$D\subset\displaystyle
\mathop{\cup}_{\sigma\in K_0, \tau\in L_0}\sigma\times\tau$.
We will use this in a moment.

First, let $A\subset \vert K\vert\times_C\vert L\vert$ be closed.
This means that for every compact 
$D\subset \vert K\vert\times\vert L\vert$, $A\cap D$ is a closed
subset of $D$.
But $\sigma\times\tau$ is a product of compact sets, hence compact
so consider the case $D=\sigma\times\tau$.
Hence $A\cap(\sigma\times\tau)$ is a closed subset of 
$\sigma\times\tau$.
By definition, $A$ is closed in the topology on 
$\vert K\vert\times\vert L\vert$ coherent for $\sigma\times\tau$.

Conversely, suppose $A$ is closed in the topology on 
$\vert K\vert\times\vert L\vert$ coherent for the $\sigma\times\tau$:
i.e. $A\cap(\sigma\times\tau)$ is closed in $\sigma\times\tau$.
If $D$ is compact, $D\subset\displaystyle
\mathop{\cup}_{\sigma\in K_0, \tau\in L_0}\sigma\times\tau$,
so $A\cap D=D\cap\bigl(\displaystyle
\mathop{\cup}_{\sigma\in K_0, \tau\in L_0}(\sigma\times\tau)
\cap A\bigr)$.
Each $(\sigma\times\tau)\cap A$ is closed in $\sigma\times\tau$
and there are only finitely many elements in the union, so
$\mathop{\cup}_{\sigma\in K_0, \tau\in L_0}(\sigma\times\tau)
\cap A$ is closed in 
$\mathop{\cup}_{\sigma\in K_0, \tau\in L_0}\sigma\times\tau$.
But $\mathop{\cup}_{\sigma\in K_0, \tau\in L_0}\sigma\times\tau$
is a compact subset of a Hausdorff space, hence closed in
$\vert K\vert\times\vert L\vert$.
Hence $\mathop{\cup}_{\sigma\in K_0, \tau\in L_0}(\sigma\times\tau)
\cap A$ is closed in $\vert K\vert\times\vert L\vert$.
It follows that $A\cap D$ is closed in $D$ and therefore 
$A$ is closed in $\vert K\vert\times_C\vert L\vert$.




\end

lemma 2.5 p.10