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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

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It sounds like you might be missing the general lifting theorem for covering maps. Here's the statement.

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.

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