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?


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.


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