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This claim is from Ulrich Gortz and Torsten Wedhorn's Algebraic Geometry I. First I think there are some errors that (ii) and (iii) should be

For every topological space $Y$ and every continuous map $f : Y\to X$ its graph $\{(y,f(y)); y\in Y\}$ is closed in $Y\times X$.

For every topological space $Y$ and any two continuous maps $f, g : Y\to X$ the kernel $\{y\in Y ; f(y) = g(y)\}$ is closed in $Y$.

But even for this, I wonder if this is correct? For example, I know that if $X$ is Hausdorff, then for every topological space $Y$ and every continuous map $f : Y\to X$ its graph $\{(y,f(y)); y\in Y\}$ is closed in $Y\times X$, but is the converse true?

Similarly, I wonder if for every topological space $Y$ and any two continuous maps $f, g : Y\to X$ the kernel $\{y\in Y ; f(y) = g(y)\}$ is closed in $Y$ we can conclude that $X$ is Hausdorff?

6666
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  • $X,Y$ should also be interchanged in $(iii)$ so that the indicated 'kernel' makes sense as a subset of the domain. – Oliver Kayende Nov 15 '19 at 17:11
  • @Mirko are you sure? I think William's answer works perfectly well with my correction – 6666 Nov 15 '19 at 20:46
  • @Mirko the Hausdorff space needs to be the image for the graph statement do $Y\times X$ is correct. And the kernel is a subspace of $Y$ indeed. – Henno Brandsma Nov 15 '19 at 22:24

1 Answers1

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Hausdorff implies (i).
If (x,y) not in the diagonal, exists open U,V that separate x,y.
Since U,V are disjoint, open U×V misses the diagonal.
Thusly the complement of the diagonal is open.

(i) implies (iii).
f×g:Y×Y -> X×X, (x,y) -> (f(x), g(x)) is continuous.
{ y : f(y) = g(y) } = (f×g)$^{-1}$({ (x,y) : x = y }
is the inverse image of a closed set by a continuous function, hence closed.

(iii) implies (ii). (ii) is a special case of (ii).
(ii) implies (i). (i) is a special case of (i).
(i) implies Hausdorff.
Proof is left for the enjoyment of the diligent reader,

William Elliot
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