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I'm just curious to make sure that I'm not missing something.

The CW complex for $S^1 \vee S^2$ consists of

one 0-cell (the fixed point),

one 1-cell ($S^1$) and

one 2-cell ($S^2)$?

Hence, if I were to compute the homology groups of $S^1 \vee S^2$, the $C_n$ chain would be given by $$0 \longrightarrow \mathbb{Z} \longrightarrow \mathbb{Z} \longrightarrow \mathbb{Z} \longrightarrow 0.$$ Is this correct?

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Yes, that sequence will work, you just need to describe what each map is doing (each one should be trivial). To be more specific though, the 1 and 2 cells are 1 and 2 disks, respectively, but attached to a point (thus creating an $S^1$ and $S^2$). Also, you may be interested in

The homology of wedge sum

which shows that, for 'good pairs', the reduced homology of the wedge is the sum of the reduced homologies.

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