1
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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