I understood the formal definition of isomorphism and homomorphism.But I am not to make out difference between them.I read it here, but they say that homomorphism loses information while isomorphism preserves it.Please explain what do they mean by losing information and what do they(isomorphism and homomorphism) signify.
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1Why do you think $f(3+7) = f(3)+f(7)$ is not true? In $\mathbb Z_2$ you have $1+1=0$. – Christoph Oct 13 '15 at 11:40
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Ok I got it my silly mistake. – Siddharth Oct 13 '15 at 11:43
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Yeah that's fine, but what do they mean by losing information I don't understand.Conceptually I am not getting isomorphism and homomorphism. – Siddharth Oct 13 '15 at 11:46
3 Answers
When people say that a homomorphism "loses information" while an isomorphism does not, what they mean is that an isomorphism is something that you can undo. It depends on how you define an isomorphism, but my preferred definition is the following.
Definition: And isomorphism $f : A \to B$ is a (homo)morphism such that there exists a (homo)morphism $g : B \to A$ such that $f \circ g : B \to B$ and $g \circ f : A \to A$ are both the identity.
This may not be the same as your definition, but I encourage you to think about why these are actually going to be the same.
One should note that every isomorphism is in fact a homomorphism, and so these are a restricted class of homomorphisms.
The key though is that the existence of the inverse morphism $f^{-1} = g: B \to A$ allows us to "undo" what $f$ does; that is, for every $a \in A$, if we map this to $f(a) \in B$, we can recover $a$ by applying $g$.
Now, let's look by contrast at the morphism described above. The problem is that we have $f(2) = f(0) = f(100) = f(-22) = f(12345678) = 0$ etc. and so starting with the image in $B$, we can not uniquely recover the pre-image in $A$. So we have lost information.
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Isomorphism are 1 to 1 and onto and so are invertible. Every isomorphism is a Homomorphism since they have to operation preserving properties that define a Homomorphism.
Losing information means the follwing :
Suppose $f(x)=f(y)=z$ , $x\ne y$.
In this case, given the value of $f(x)=z$, you cannot determine the value $x$ because there are different possible values. This is not the case, if $f$ is a monomorphism. In this case, $x$ is uniquely determined by $z$.
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