# Homology with local coefficients (question on Whitehead's proof)

I am reading [1] and trying to understand the identification of the cohomology with local coefficients $$\bar{M}$$ on a space $$X$$ and cohomology of $$C_*(\tilde{X}) \otimes_G M$$ with $$M$$ a $$\mathbb{Z}[G]$$-module. ($$\tilde{X}$$ is the universal covering of $$X$$ and $$G = \pi_1(X,x_0)$$).

I summarize the idea of the proof: $$M$$ is identified with $$\bar{M}_{x_0}$$, and a map $$p: M \otimes_G C_*(\tilde{X}) \rightarrow C_*(X; \bar{M})$$ is constructed in the following way.

For an element $$g_0 \otimes \sigma \in M \otimes_G C_*(\tilde{X})$$, $$p(g_0 \otimes \sigma) = h_{p\circ \gamma}(g_0) \cdot (p\circ \sigma)$$ where $$g = h_{p\circ \gamma}: \bar{M}_{p(x_0)} \rightarrow \bar{M}_{p(y)}$$ where $$\gamma$$ is a path connecting $$x_0$$ and $$y = \sigma(e_0)$$.

Then it is claimed that the it is not difficult to see map is an isomorphism. However, I am having issues convincing myself that the map is surjective. Namely, if $$\sigma: \Delta^n \rightarrow X$$ is a simplex with $$n \geq 2$$, then it admits a lifting to $$\tilde{X}$$. However, if $$n = 1$$, the lifting is just a path and I do not see how it can be the image of some $$1$$-simplex upstairs.

[1.] Whitehead, Elements of Homotopy Theory, pg. 278

Let $$p : Y \to X$$ be a covering map, and let $$f : Z \to X$$ be a continuous map, where $$X,Y,Z$$ are path connected and locally path connected, and let $$x \in X$$, $$y \in Y$$, and $$z \in Z$$ be such that $$f(z)=p(y)=x$$.
Question: Does a lift exists? Meaning, does there exist a continuous function $$\tilde f : Z \to Y$$ such that $$\tilde f(z)=y$$ and such that $$p \circ \tilde f = f : Z \to X$$?
Answer: the lift $$\tilde f$$ exists if and only if the image of the induced homomorphism $$f_* : \pi_1(Z,z) \to \pi_1(X,x)$$ is a subgroup of the image of the induced monomorphism $$p_* : \pi_1(Y,y) \to \pi_1(X,x)$$.
In your case, because the fundamental group of $$\Delta^n$$ is trivial, it follows that a lift of the function $$\sigma : \Delta^n \to X$$ does indeed exist.