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I've recently been teaching myself algebraic topology from Croom's Basic Concepts of Algebraic Topology, and I'm really lost on a problem concerning the Euler characteristic of a pseudomanifold. The problem is as follows:

Prove that if $K$ is a 2-pseudomanifold, then $\chi(K)\leq 2$, where $\chi(K)$ is the Euler characteristic of $K$.

I can see how this intuitively makes sense, but I really am not sure of where to start or how to write a formal proof of this. Any help is appreciated!

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    Do you know anything about the homology of pseudomanifolds? – 2017-01-02
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    @EricWofsey, I do know that if a pseudomanifold is orientable, then its n-homology group is non-trivial. Other than that, I haven't come across anything else. Is there something obvious that I'm missing? – 2017-01-02
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    If an $n$-dimensional pseudomanifold is orientable, do you know exactly what its $n$th homology group is? And what about if it's not orientable? – 2017-01-02
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    @EricWofsey, if an $n$-dimensional pseudomanifold is orientable, then $H_{n}(K)\simeq \mathbb{Z}$, right? And if it is not orientable, then $H_{n}(K)=0$? – 2017-01-02
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    OK, so now try computing $\chi(K)$ using the homology of $K$, if $n=2$. – 2017-01-02
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    How exactly would you go about that? I think not having an abstract pseudomanifold rather than a concrete example is messing with my head. – 2017-01-02

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Hint: You can compute $\chi(K)$ as the alternating sum of the ranks of the homology groups of $K$, which in this case go up to only dimension $2$. So $\chi(K)=\operatorname{rank} H_0(K)-\operatorname{rank} H_1(K)+\operatorname{rank} H_2(K)$. What possible values can each of the numbers appearing in this sum take? (Don't worry if you don't know exactly what all of them are; remember that you only need to prove that $\chi(K)\leq 2$.)

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    Thanks! That makes more sense now! – 2017-01-02