Constant sheaves and morphisms between them

Question

All my sheaves are sheaves of $k$ modules, where $k$ is a commutative ring with unity. Let $X$ be a topological space, which is locally connected and connected. Let $\phi$ from $k_X$ to $k_X$ be a morphism of sheaves, where $k_X$ is the constant sheaf on $X$, that is the sheaf of all locally constant functions taking values in $k$. Then I want to show that $\phi(U)$ is multiplication by a fixed constant (say $b$) for any open subset $U$ of $X$. Please help me.

Answer 1

Correct me if I am wrong

First consider the case when $U$ is connected. Then it is obvious (the $k$-morphisms from $k$ to $k$ are just multiplications by scalar). Now let $U= \bigcup_{i \in I} A_i$ where all $A_i$ are connected. On each of the $A_i \ $ $\phi(U)$ acts by multiplication. Now look at the element $1_X$ in $k_X(X)$. It is mapped by $\phi(X)$ to the function $a \cdot 1_X$. But $\phi$ commutes with restrictions. Thus the scalar on each $A_i$ must be the same and equal to $a$.

Answer 2

As shown by user68061, stalks are not necessary, but you could do it this way: all the stalks are $k$ because $X$ is locally connected, $\phi$ is the multiplication by some constant at every point, so you have a function assigning to every point that constant. This function is easily seen locally constant, hence globally constant because $X$ is connected, and this is enough.