For $f:X\rightarrow Y$ continuous map, and $Z$ a closed subset of $Y$. I want to constract the natural isomorphism $f^{-1}k_{YZ}\rightarrow k_{Xf^{-1}Z}$. Where $f^{-1}k_{YZ}$ is the pullback of the sheaf $k_{YZ}:=(\iota_{Z*}\circ\iota_Z^{-1})k_Y$ and $k_Y$ is the constant sheaf with stalk $k$. Similarly $k_{Xf^{-1}Z}:=(\iota_{f^{-1}Z*}\circ\iota_{f^{-1}Z}^{-1})k_X$ ($\iota$, is the inclusion).
The fact that both sheaves agree on the stalks is straightforward, but I am having difficulties trying to find the natural map.
Any hints?
For convenience let $W$ denote $f^{-1}(Z)$.
First we show that $f^{-1}(k_Y)\cong k_X$. Let $T$ be the topological space with one point, and let $g:Y\rightarrow T$ be the unique continuous map. Then $k_Y$ is the pull-back $g^{-1}(k_T)$. So $f^{-1}k_Y\cong f^{-1}g^{-1}(k_T)\cong (g\circ f)^{-1}(k_T)$ is the constant sheaf on $X$.
Now by the adjunction between the pull-back and the push-forward we know that to have a morphism $f^{-1}(k_{YZ})\rightarrow k_{X,f^{-1}Z}$ is the same as to have a morphism $\iota_{Z*}\circ\iota_Z^{-1}(k_Y)\rightarrow f_*\circ\iota_{W*}\iota^{-1}_W\circ f^{-1}(k_Y)$. But $f\circ\iota_W=\iota_Z\circ f$, so the right-hand side is $\iota_{Z*}\circ f_*\circ f^{-1}\circ\iota_Z^{-1}(k_Y)$. Hence from the canonical adjunction morphism $\operatorname{id}\rightarrow f_*\circ f^{-1}$ we obtain the canonical morphism $\iota_{Z*}\circ\iota_Z^{-1}\rightarrow \iota_{Z*}\circ f_*\circ f^{-1}\circ\iota_Z^{-1}$.
Hope this helps.