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Let $A$ be a finite Boolean algebra. If I define a monomorphism (i.e. an injective homomorphism) from $A$ to another finite Boolean algebra $B$ of the same similarity type. Is this monomorphism an isomorphism?

I am tempted to think that yes, for two Boolean algebras which have the same number of elements are isomorphic. But I would like a second opinion to be sure...

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    $\begingroup$ What do you mean by "similarity type"? $\endgroup$ – Eric Wofsey Jun 2 '16 at 16:19
  • $\begingroup$ @EricWofsey. I mean same signature. $\endgroup$ – user60264 Jun 2 '16 at 16:30
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    $\begingroup$ What is the "signature" of a Boolean algebra? $\endgroup$ – Eric Wofsey Jun 2 '16 at 16:40
  • $\begingroup$ @EricWofsey. Boolean algebras can have many different signatures (a term rather found in logic) or similarity types (rather found in universal algebra). For instance, a Boolean algebra can be defined in terms of join and complement. Or, join, meet, and complement, etc... Typically, a Boolean algebra would be written $(A,\vee,\wedge,-)$ for instance. $(\vee,\wedge,-)$ is the type or signature of $A$. $\endgroup$ – user60264 Jun 2 '16 at 16:55
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No, this isn't true, because the homomorphism need not be surjective. For instance, $f:X\to Y$ is any surjective map of finite sets, then the inverse image map $f^{-1}:P(Y)\to P(X)$ is an injective homomorphism of finite Boolean algebras. But $P(Y)$ and $P(X)$ have different cardinalities if $X$ and $Y$ have different cardinalities, in which case $f^{-1}$ is not an isomorphism.

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    $\begingroup$ +1. A simpler counterexample: there's a monomorphism from the 2-element Boolean algebra to any Boolean algebra (send top to top and bottom to bottom). $\endgroup$ – Noah Schweber Jun 2 '16 at 17:07
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    $\begingroup$ @NoahSchweber: To any Boolean algebra except the Boolean algebra with one element! $\endgroup$ – Eric Wofsey Jun 2 '16 at 17:08
  • $\begingroup$ Aaargh . . . :P Quite right, quite right (and here I am with a universal algebra textbook even! no excuse. sigh). $\endgroup$ – Noah Schweber Jun 2 '16 at 17:12
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    $\begingroup$ @student: Yes, because any injection between two finite sets of the same cardinality is a bijection. $\endgroup$ – Eric Wofsey Jun 2 '16 at 17:25
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    $\begingroup$ @student Note that this crucially assumes that the Boolean algebras in question are finite - that said, it also holds for all types of structures, not just Boolean algebras, as long as they are finite. $\endgroup$ – Noah Schweber Jun 2 '16 at 18:17

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