Is the mapping cylinder of a map between two locally compact Hausdorff spaces locally compact? In Spanier's book "Algebraic Topology" on page 97 Sec 2.8, he proves a theorem about construction of fibration from cofibration.

THEOREM. Let $f:X' \rightarrow X$ be a cofibration, where $X'$ and $X$ are locally compact Hausdorff spaces, and let $Y$ be any space. Then the map $p: Y^X \rightarrow Y^{X'}$ defined by $p(g) = g \circ f$ is a fibration.

In the proof he claims that:

Because $X'$ and $X$ are locally compact Hausdorff spaces, so is $\bar{X}.$

Here $\bar{X}$ is the mapping cylinder of $f$ defined by the quotient space of $(X' \times [0,1]) \vee (X\times 0)$ under the equivalence $(x',0)\thicksim (f(x'),0)$.
The claim above is clear because here the cofibration is actually a closed inclusion map.
My question is: in general, is the mapping cylinder of a map between two locally compact Hausdorff spaces still locally compact?

No.  For instance, consider the unique map $\mathbb{N}\to \{*\}$.  Its mapping cylinder is the cone on $\mathbb{N}$, which is not locally compact at the cone point.  (For instance, for sequence of numbers $\epsilon_n>0$, the sequence $(n,\epsilon_n)$ has no accumulation point, and every neighborhood of the cone point contains such a sequence.)
A sufficient general condition for the mapping cylinder to be locally compact is that $f$ is proper.  Indeed, if $f$ is proper, then given a point $(x,0)\in X\times\{0\}$, you can pick a compact neighborhood $K$ of $x$ and then $f^{-1}(K)\times[0,1]\cup K\times\{0\}$ gives a compact neighborhood of $(x,0)$ in the mapping cone.
