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I am readying Hatcher page 178 where he tries to give the idea of cohomology in the case where our space $X$ is a graph. Now in the 4th paragraph from the top he says this:

The cohomology group $H^1(X,G) = \Delta^1(X;G)/\textrm{Im} \delta$ will be trivial iff the equation $\delta \varphi = \psi$ has a solution $\varphi \in \Delta^0 (X;G)$ for each $\psi \in \Delta^1 (X;G)$. Solving this equation means deciding whether specifying the change in $\varphi$ across each edge of $X$ determines an actual function $\varphi \in \Delta^0(X;G)$.

What does he mean by that last sentence in bold in the paragraph above?

Thanks.

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If $e = [v_0,v_1]$ is an (oriented) edge, then $\delta\varphi[v_0,v_1] = \varphi(v_1) - \varphi(v_0)$ can be interpreted as the change of $\varphi$ on the edge $e$ (hence the notation reminiscent of the derivative). The question if there is $\varphi$ such that $\delta\varphi = \psi$ asks if $\psi$ is the change $\delta \varphi$ of a function $\varphi$ on the vertices. The non-vanishing of $H^1(X,G)$ means precisely that there are functions $\psi$ on the edges that do not arise as change $\delta \varphi$ of a function $\varphi$ on the vertices. –  commenter Oct 31 '12 at 10:02

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up vote 10 down vote accepted

I like to think of this as follows. If you're hiking in the mountains and you keep track of your changes in elevation, then if you ever return to the same spot your total change in elevation will be $0$. So, $H^1$ detects whether it's possible to have a sensible notion of "local change in elevation" such that you don't necessarily return to $0$ when you come around to your starting point. A great illustration of this possibility is the "never-ending staircase";

never-ending staircase

note that this is only possible because we're not required to say what happens with our elevation anywhere in the center hole. (That is, $H^1(S^1)\not= 0$ but $H^1(D^2)=0$.)

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This is a really great interpretation of (lower) cohomology! –  Miha Habič Oct 31 '12 at 17:15

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