# Deck or Monodromy actions to define homology with local coefficients

when we have a covering $$p:Y\to X$$ choose $$x\in X$$ and we have two actions on the fiber $$p^{-1}(x)$$:

1) The monodromy action of $$\pi_1(X,x)$$ defined as follows: $$[\gamma].\tilde x=\tilde \gamma (1)$$ that is; the action gives the endpoint of the unique lift of $$\gamma$$ starting at $$\tilde x$$.

2)The Deck group action of the group of Deck transformations $$Deck(Y,X)$$ on $$p^{-1}(x)$$ defined as follows: $$\phi. \tilde x=\phi(\tilde x)$$

When $$Y=\tilde X$$ is the universal covering we have that the two groups $$Deck(\tilde X,X)$$ and $$\pi_1(X,x)$$ are isomorphic.

Now my problem is the following : when defining homology of $$X$$ with local coefficients in a $$\mathbb Z[\pi_1(X)]-$$module $$A$$, many authors say that $$\pi_1(X,x)$$ acts on $$\tilde X$$ by Deck transformations and set $$C_n(X,A):=C_n(\tilde X,\mathbb Z)\otimes A$$

so first why they don't say $$\pi_1(X)$$ acts on $$\tilde X$$ by monodromy action as defined above instead of saying that it acts by Deck transformatioins and second why taking the universal cover when they could take any cover of $$X$$ and we still have monodromy and Deck actions.

• For you second question: The action will fail to be proper so the quotient might not be Hausdorff. For the first question: You get the same answer as you noted. Commented Jul 15, 2019 at 14:48
• @MoisheKohan where does the quotient $\tilde X/\pi_1(X,x)$ appear in this construction ? Commented Jul 15, 2019 at 15:46
• It appears as the base of a fibration. But what I wrote refers to a different quotient. As an extreme case, take $X=\tilde{X}$. Commented Jul 15, 2019 at 15:56
• @MoisheKohan In Hatcher page 72 : the action of the Deck transformation group is properly discontinuous. Commented Jul 15, 2019 at 16:08
• Sure, but you are proposing to use the action of $\pi_1(X)$ which is not the same as the deck group, unless you have the universal covering space. Incidentally, your monodromy does not even define an action of $\pi_1(X)$ unless you have a regular covering. I suggest that you work out some examples when $\tilde{X}$ is not simply connected and see what you get. Commented Jul 15, 2019 at 16:17

Given a covering map $$p:Y\to X$$, the Deck transformations group $$Deck(Y,X)$$ acts on $$Y$$ and this action restricts to an action on each fiber $$p^{-1}(x)$$. The monodromy action however is an action of $$\pi_1(X,x)$$ on the fiber $$p^{-1}(x)$$ and this action need not be defined on the whole space $$Y$$ unless the covering $$p:Y \to X$$ is a regular covering.
When the covering $$Y$$ is the universal covering $$\tilde X$$, we have that $$\pi_1(X,x_0)$$ is isomorphic to $$Deck(\tilde X,X)$$ and that's why we talk about the action of $$\pi_1(X,x)$$ on $$\tilde X$$ by Deck transformations.