Write $\mathbb F_2=\mathbb Z/2$. A double covering has a <deck transformation> $\tau$ that exchanges the two points in every fibre. Every singular simplex $\sigma:\Delta^q\to Y$ has exactly two lifts $\widetilde\sigma$ and $\tau\widetilde\sigma$. Define the mod-two <transfer chain map of a double covering> by
$$
T(\sigma)=\widetilde\sigma+\tau\widetilde\sigma,
$$
and let $f_\#$ send a simplex of $X$ to its composite with $f$. Uniqueness of lifted faces shows that $T$ commutes with the <boundary operator>, so both maps are <chain map>[chain maps].

For each base simplex, its two lifts span a copy of $\mathbb F_2^2$. On this summand, $T$ is $a\mapsto(a,a)$ and $f_\#$ is $(a,b)\mapsto a+b$. Hence the sequence is exact in every degree:
$$
\boxed{0\longrightarrow C_*(Y;\mathbb F_2)
\xrightarrow{T}C_*(X;\mathbb F_2)
\xrightarrow{f_\#}C_*(Y;\mathbb F_2)
\longrightarrow0.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 114 / 2 / a / Solution

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The boundary operator of a <chain complex> is its degree-minus-one differential $\partial:C_i\to C_{i-1}$. The identity $\partial^2=0$ says that every <chain boundary> is a <chain cycle>.

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 Incoming links: Boundary operator