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A ring $R$ is called to be 'Hopfian' if every ring homomorphism of $R$ onto $R$ is an automorphism of $R$

Question: Given $R$, $S$, $T$, Hopfian rings and $$R \times S \cong T \times S$$ implies that there exists an isomorphism $$R \cong T$$

My approach:

Let $X1, X2, Y \in C$. According to product rule in Category theory for every object Y and pair of morphisms

$$f1 : Y \to X1, \space\space f2 : Y → X2$$

there exists a unique morphism $$f : Y → X1 × X2$$ where $f = <f_1, f_2>$

Given $f: R \times S \to T \times S$, therefore $f = <f_1, f_2>$ where $f_1: R \times S \to T$ and $f_2: R \times S \to S$. Since $S$ is Hopfian, therefore $f_2$ is an isomorphic function. I wonder if isomorphism of $f$ does implies isomorphism of $f_1$ and $f_2$. Also please let me know if this is the correct approach or should I look at it in some other way.

P.S: This is my first question on mathexchange, please help me correct any mistakes which I might have made over here.

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  • $\begingroup$ Do we consider ring homomorphisms to preserve the multiplicative identity? How is $f_2:R\times S\to S$ an iso? $\endgroup$ – Berci Dec 10 '18 at 8:27
  • $\begingroup$ Yes, ring homomorphism preserves multiplicative identity in this case. $f_2: R \times S \to S$ where $f_2(r,s) \to s$ is a bijective mapping since S is given to be hopfian object. $\endgroup$ – forcehandler Dec 10 '18 at 8:50

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