- #1

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The next step is what I have a question about. In this case would it be better to show that ##\gamma (b,a) = (a,b)## is the inverse of ##\phi##, or show explicitly that ##\phi## is injective and surjective?

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- I
- Thread starter Mr Davis 97
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- #1

- 1,462

- 44

The next step is what I have a question about. In this case would it be better to show that ##\gamma (b,a) = (a,b)## is the inverse of ##\phi##, or show explicitly that ##\phi## is injective and surjective?

- #2

fresh_42

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Both would be o.k., but it's so obvious, that you don't really need a proof. For the version with ##\gamma## don't forget, that both have to be proven: ##\gamma \phi = \operatorname{id}_{A \times B}## and ##\phi \gamma = \operatorname{id}_{B \times A}##. However, surjectivity is immediately clear and so is injectivity.

The next step is what I have a question about. In this case would it be better to show that ##\gamma (b,a) = (a,b)## is the inverse of ##\phi##, or show explicitly that ##\phi## is injective and surjective?

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