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Lemma 24: Fix basepoints $x,y,y' \in X,Y,Y'$ as usual. There is a lift $f':X'\rightarrow Y'$ of $f:X\rightarrow Y$ such that $f(x)=y'$ if and only if $f_*\pi_1(X,x) \subseteq \pi_1(Y',y')$.

Furthermore, if the lift exists, it is unique.

Lemma 25: Given a choice of $y'$, there exists an elevation $f':X'\rightarrow Y'$ of $f:X\rightarrow Y$ at $y'$. Furthermore, $f'$ is unique in the sense that if $\bar{f}:\bar{X}\rightarrow Y'$ is another elevation of $f$ and $y'$ then there is a homeomorphism $X'\rightarrow \bar{X}$ and the diagram commutes.

Proof: Let $(X',x')\rightarrow (X,x)$ be defined by $\pi_1(X',x')=f^{-1}_*\pi_1(Y',y')$. By Lemma 24, the composition $(X',x')\rightarrow (\bar{X},\bar{x})\rightarrow (Y,y)$

lifts at $y'$, call this lift $f'$. Suppose Then $\pi_1(X',x')\subseteq \pi_1(\bar{X}\bar{x})$. But by Lemma 24, $f_*\pi_1(\bar{X},\bar{x})\subseteq \pi_1(Y',y')$. This implies $\pi_1(X',x') =\pi_1(\bar{X},\bar{x})$.

Another, more categorical construction uses the fibre product. The fibre product $\hat{X}$ is defined by $\hat{X} =\{(\xi,\eta)\in X\times Y'|f(\xi)=\sigma(\eta) \}$.

There are obvious maps $\hat{X}\stackrel{\hat{f}}{\rightarrow} Y'$ $\hat{X}\stackrel{f}{\rightarrow} X$

given by forgetting factors.

Exercise: If $\sigma$ is a covering map then $\rho$ is a covering map.

Lemma 26: Fix $x\in X$. Let $y=f(x)\in Y$ and let $\sigma(y')=y$ for $y'\in Y$. Let $x'=(x,y')\in \hat{X}$, and let $X\subseteq\hat{X}$ be the connected component containing $x'$. Then $f'=\hat{f}|_{x'}:X'\rightarrow Y'$ is an elevation of $f$ at $y'$, and every elevation of $f$ arises in this way.

Proof: To prove that $f'$ is an elevation we just observe that $p_*\pi_1(X',x')=f_*^{-1}\sigma_*\pi_1(Y,'y')$. Now suppose is an elevation. Then $\bar{X}\rightarrow \hat{X}=\{(\xi,\eta)|f(\xi)=\sigma(\eta) \}$, with $\xi\mapsto(\tau(\xi),\bar{f}(\xi))$.

The covering map $\tau$ factors trhough $\bar{X}\rightarrow\hat{X}$, and so $\bar{X}\mapsto\hat{X}$ is a covering map. Because $\bar{X}$ is an elevation, $\bar{X}\mapsto\hat{X}$ is a homeomorphism onto its image, a connected component of $\hat{X}$.

What has this got to do with graphs of spaces/groups?

Let $Y$ be a vector space, and let $\partial: E\rightarrow Y$ be an edge map. Define $X$ to be the mapping cylinder $X =(E\times[0,1]) \sqcup Y/\sim ;\;\; (x,1)\sim\partial(x)$ $X$ comes with a map $d:X\rightarrow Y$ such that $d|_Y=id_Y$ and $d(x,t)=\partial(x)$. This is an inclusion $\iota:Y\rightarrow X$, and $d\circ\iota =id_Y$. Let $\sigma:Y'\rightarrow Y$ be a covering map. Let $\hat{X}$ be the fibre product. There’s a map $\iota':Y'\rightarrow \hat{X}$; $\eta\mapsto (\iota\circ\sigma (\eta),\eta)$. Clearly, $d'\circ\iota'=id_{Y'}$.

Therefore, $\iota'$ is an injection. It’s easy to see that $\iota'$ induces a bijection at the level $\pi_0$.

Lemma 27: Any covering space $X'\rightarrow X$ arises as the fibre product of a covering map $Y'\rightarrow Y$.

Proof: Let $\tau:X'\rightarrow X$ be a covering map let $Y'$ be the fibre product of $\tau$ and $\iota$. $Y'=\{(\xi,\eta)\in X'\times Y |\tau(\xi)=\iota(\eta)\}$

Define $d':X'\rightarrow Y'$ by $\xi\mapsto (\xi,d\circ\tau(\xi))$. As before, $\sigma$ is a covering map.