TITLE: Non-injective monomorphisms
QUESTION [8 upvotes]: I am reading Borceux, vol. 1, I found this example at page 27:
we consider the category whose object are the pairs $\langle X,x\rangle$ where $X$ is a topological space and $x$ a point of $X$ (base point). In this category, a morphism $f:\langle X,x\rangle\longrightarrow\langle Y,y\rangle$ is a continuous mapping $f:X\longrightarrow Y$ which preserves base points. Let us consider the projection $\pi$ of the circular helix $\mathcal{H}$ on the circle $\mathcal{S}^1$,
$$\pi:\langle \mathcal{H},h\rangle\longrightarrow\langle\mathcal{S}^1,s\rangle$$
with $h\in\mathcal{H}$ and $s=\pi(h)$. If $f:\langle X,x\rangle\longrightarrow\langle\mathcal{S}^1,s\rangle$ is a morphism in our category which admits a lifting
$$g:\langle X,x\rangle\longrightarrow\langle\mathcal{H},h\rangle$$
through the projection $\pi$, that lifting is necessarily unique (see Spanier, pag 67), and so $\pi$ is a (non-injective) monomorphism...
This is pretty clear to me, except for a single detail: why do we need to consider base points (and morphisms preserving them)? Is there a problem if we simply consider the category $\operatorname{\mathbf{Top}}$ of topological spaces and continuous mappings?

REPLY [7 votes]: Borceux's example is incorrect. Here is a general observation.

Let $(C, U : C \to \text{Set})$ be a concrete category such that the forgetful functor $U$ to sets has a left adjoint. Then monomorphisms are injective (meaning that $U$ preserves monomorphisms). 

This covers most familiar examples of concrete categories, including topological spaces, groups, rings, etc. In particular, it includes based topological spaces! Here the left adjoint takes a set $X$ to the pointed space given by $X$ with the discrete topology plus a new basepoint. 
You can use this to show that Borceux's map, which amounts to the exponential map $\exp : (\mathbb{R}, 0) \to (S^1, 1)$, is not a monomorphism. If we let $2 = 1 + 1$ be a point and a basepoint, then the map $2 \to S^1$ sending everything to the basepoint has many lifts, given by sending the non-basepoint to any lift of the basepoint in $\mathbb{R}$. 
This example can be fixed by restricting further to the category of path connected (and maybe locally path connected, to be on the safe side) based spaces; in this category any covering map is a monomorphism because unique lifting actually holds here. Here the forgetful functor no longer has a left adjoint, so the above no longer applies.