Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Here are some exercises from Rotman's book (homological algebra).

$1$. Show that $\textbf{Top}$ has no zero object.

Let $X$ be a zero object in $\textbf{Top}$ then as X is initial and $\emptyset$ is a topological space then there is a unique continuous map $f: X \rightarrow \emptyset$. Therefore $X$ is the empty set, since $f$ is the empty function (which is continuous).

On the other hand, since $X$ is terminal choose a non-empty topological space $Z$. Thus there is a unique continuous map $f: Z \rightarrow \emptyset$ but there is no such function since $Z \neq \emptyset$.

$2$. Prove that the zero ring is not an initial object in the category of commutative rings.

Can we simply say: assume there is a ring homomorphism $f: 0 \rightarrow \mathbb{Z}$ where $0$ denotes, by abuse of notation, the zero ring. On one hand $f(0)=1$ because $f$ preserves the unity and $0=1$ in the zero ring. But on the other hand $f(0)=0$ so $0=1$ in $\mathbb{Z}$ which is absurd.

Is this OK? thanks for your time/help.

share|improve this question
In the second paragraph of 1., you mean "there is a unique continuous map $f: Z\rightarrow \emptyset$." –  Alex Kruckman Dec 21 '12 at 3:44
@Alex Kruckman: right, thanks! –  user10 Dec 21 '12 at 3:47

1 Answer 1

up vote 4 down vote accepted

Your arguments for 1 are perfectly fine.

As for 2, it depends on the precise definition of the notion of morphisms in the category. It seems that the category you work with demands the morphisms to be ring homomorphism between unital (commutative) rings that further preserve the unit. If that is the case then your argument is correct.

share|improve this answer
yes, the homomorphism is assumed to be preserve units. Thanks! –  user10 Dec 21 '12 at 3:38
You're welcome. –  Ittay Weiss Dec 21 '12 at 3:39

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.