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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\mathcal{C}$ be the category of finite covers of a fixed base space $S$ (say, connected, locally path connected, locally simply connected. Hell, we can even assume $S$ is a manifold). Morphisms are just morphisms of covering spaces.

Suppose $X\in\mathcal{C}$ is connected, then any monomorphism (i.e., injection) from any $T\in\mathcal{C}$ to $X$ must necessarily be a covering map of degree 1, and hence an isomorphism.

Now suppose we have two maps $u_1, u_2 : X\rightarrow Y$, then we may construct the equalizer $E$ of $u_1, u_2$, and get a diagram $$E\rightarrow X\stackrel{\longrightarrow}{\longrightarrow}Y$$

Since the equalizer of the two maps can be equivalently expressed as $(X\times_Y X)\times_{X\times X} X$ (where $X\rightarrow X\times X$ is the diagonal and $X\times_Y X\rightarrow X\times X$ is given by the projections $p_1,p_2 : X\times_Y X\rightarrow X$ corresponding to the maps $u_1,u_2 : X\rightarrow Y$), and since fiber products exist in $\mathcal{C}$, we find that $E$ is also an object of $\mathcal{C}$. (this is from Murre's introduction to grothendieck's theory of the fundamental group).

Hence, the diagram above is a diagram in $\mathcal{C}$, and hence the first map $E\rightarrow X$ must be a monomorphism (since $E$ is an equalizer), and hence an isomorphism (since $X$ is connected). Thus, $u_1 = u_2$.

What's wrong with this proof?

share|cite|improve this question
up vote 3 down vote accepted

The problem is in your second paragraph. If $T \to X$ is monomorphism, then either $T$ is empty or $T \to X$ is an isomorphism.

share|cite|improve this answer

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.