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I am learning about simplicial homology from Hatcher's Algebraic Topology book. He has examples of some simplicial homology computations for familiar/simple spaces (the torus, circle, and $\mathbb{R} \mathbb{P} ^2$). However, I am interested in doing a computation which involves finding a triangulation of something "more complicated," such as $S^1 \vee S^2$, and over $\mathbb{Q}$ instead of $\mathbb{Z}$ (what does the triangulation of $S^1 \vee S^2$ even look like?). That is, what is a triangulation of $S^1 \vee S^2$ and how can we compute the simplicial homology of $S^1 \vee S^2$ over $\mathbb{Q}$? (If this has been done before, I would appreciate a reference as I have tried searching for this online, and I apologize in advance if this is a duplicate.)

Compact
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    Generally, if you want to find the triangulation of $M_1 \lor M_2$, you find the triangulation of $M_1$ and $M_2$ individually and then "combine" them by giving the same label to the distinguished 0-simplex of each of their triangulations. – Mark Saving Apr 20 '21 at 20:08

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As long as one knows a triangulation of $\mathbb{S}^{1}$ and a triangulation of $\mathbb{S}^{2}$, one knows a triangulation of the wedge sum (as is probably clear) where one joins both spaces at the point [0]. $\mathbb{S}^{2}$ is homeomorphic to the boundary of a 3-simplex, by embedding the boundary of the 3-simplex in such a way that one encloses the origin in $\mathbb{R}^{3}$, and doing $x \mapsto \frac{x}{|x|}$. It is easy to check this is a bijective and continuous map of compact Hausdorff spaces, and thus a homeomorphism. Similar logic works for $\mathbb{S}^{1}$, and indeed for all higher dimensional spheres. This gets you free triangulations from the n-simplex.

Computing homology with rational coefficients can be done by taking the free $\mathbb{Q}$ vector space on the simplices in each dimension of this triangulation and computing simplicial homology as you would with $\mathbb{Z}$. As you might notice, this is the same operation as tensoring the standard $\mathbb{Z}$-coefficient simplicial chain complex with $\mathbb{Q}$, and in general this is how one computes any homology with coefficients. $\mathbb{Q}$ is flat over $\mathbb{Z}$ and a little work with flatness will show you that the computation of these homology groups agrees with $H_{n}(X; \mathbb{Z})\otimes_\mathbb{Z} \mathbb{Q}$. This does not work at finite fields, as these are not flat over $\mathbb{Z}$, and one usually gets an extra $\operatorname{Tor}$ term showing up in $H_{n}(X;\mathbb{F}_{p})$; this is dictated by something called the Universal Coefficient Theorem. This theorem is mostly likely the reference you are looking for, and is both easily searchable online and in Hatcher!

Aragogh
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  • You'll also learn about cellular homology later on in Hatcher, which is a much more tractable way to compute homology of CW-complexes, which spheres are the prototypical example of. – Aragogh Apr 20 '21 at 20:23
  • Does the final result/computation result with direct sums of copies of $\mathbb{Z}$ then? The person who asked the question probably understands this reasoning, but I was curious in your explanation not knowing about flats and tensors. – Dominated Convergence Theorem Apr 20 '21 at 20:36
  • Well the final simplicial homology of $\mathbb{S}^{1} \vee \mathbb{S}^{2}$ is one copy of $\mathbb{Z}$ in degrees 0,1,2, you can check this directly using the triangulation I gave or another one of your choice. The tensor products stuff was only for doing computations over $\mathbb{Q}$. – Aragogh Apr 20 '21 at 21:10