how many circles is a connected graph homotopically equivalent to?

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I'm having troubling understanding the following theorem from this topology chapter:

Any connected graph is homotopy equivalent to the wedge of $k$ circles, with $k = \chi(G) − 1$

From my previous question, $\chi(G) = V - E $. But following the proof of the logic, I arrive at an different answer.

Basically, the proof says contracting a maximal tree $T$ of $G$ into a single node, and the graph reduces to a wedge of "several" circles (but didn't show how many is "several").

My reasoning based on this is that, a maximal tree of $G$ has $V - 1$ edges and $V$ nodes. After the contraction of $T$, the number of edges left is $E - (V - 1) = 1 - (V - E) = 1 - \chi(G)$. Each of these surviving edges becomes a circle. So there should be $1-\chi(G)$ circles, which is the opposite of what the theorem says.

Am I missing something obvious in the above calculation?

Note: the theorem is used in a corollary a couple of pages down the way, which made me think that I must have made a mistake above somewhere.

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You're missing nothing; this is just an error in the text. A connected graph $G$ is indeed homotopy equivalent to a wedge of $1-\chi(G)$ circles.