Projections from Product Topology are open I know a proof using a basis. I'd like to prove this using only the definition of the categorical product (without resorting to any construction). In particular, it is not assumed that the projections are the set theoretic projections.
 A: Exercise 9.I (Openness of product projections) in the book Categorical Structure of Closure Operators by D. Dikranjan and W. Tholen asks to show that projection maps in any topological category are open for any closure operator on that topological category. An implicity assumption (as evidenced in Chapter 5, specifically 5.8 and 5.10) is that closed subobjects are initial morphisms, i.e. that "topology-shifting morphisms" are dense.
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In particular, a categorical characterization of open maps is the following.
Consider a closure operator, i.e. a mapping of monomorphisms $M\hookrightarrow X$ to monomorphisms $\bar M\hookrightarrow X$ such that

*

*The closure of $M\hookrightarrow X$ factors through $\bar M\hookrightarrow X$

*$M\hookrightarrow N\hookrightarrow X$ implies $\bar M\hookrightarrow\bar N\hookrightarrow X$

*$\bar{f^{-1}M}\hookrightarrow f^{-1}\bar M\hookrightarrow X$ for any $f\colon X\to Y$ and $M\hookrightarrow Y$.

Define a morphism $f\colon X\to Y$ to be open if the inclusion $\bar{f^{-1}M}\hookrightarrow f^{-1}\bar M$ is an isomorphism for every $M\hookrightarrow Y$. In other words, pre-images by open maps by definition commute with the closure operator.
A slightly better equivalent definition can be found in Chapter III section 7 of Categorical Foundations - Special Topics in Order, Topology, Algebra, and Sheaf Theory. For this one, call a subobject $D\hookrightarrow X$ dense if $\bar D\cong X$. Then $f\colon X\to Y$ being open is equivalent to the assertion that every pullback of $f$ preserves dense subobjects under pullback.
To see that this recovers the usual notion, note that for any $O\subseteq Y$ we have $O\subseteq f^{-1}(f(O))$, whence that an open $O$ is disjoint from $\bar{X\setminus f^{-1}(f(O))}=\bar{f^{-1}(Y\setminus f(O))}$. If $f$ commutes with closure, then $O$ is disjoint from $f^{-1}(\bar{Y\setminus f(O)})$, which means $f(O)$ is disjoint from $\bar{Y\setminus f(O)}$, and hence that $f(O)$ is open. Conversely, let $O=X\setminus\bar{f^{-1} M}$, which is an open set disjoint from $\bar{f^{-1}(M)}\supseteq f^{-1}(M)$, and hence $f(O)$ is disjoint from $M$. Then $f$ being open implies $f(O)$ is open and so also disjoint from $\bar M$, whence $f^{-1}(\bar M)$ is disjoint from $f^{-1}(f(O))\supseteq O=X\setminus\bar{f^{-1} M}$. Thus we have a reverse inclusion $f^{-1}\bar M\subseteq\bar{f^{-1}M}$, and the desired equality.
Although understudied, this is not a useless notion: in the context of algebraic geometry, the morphisms open in this sense are the flat ones (literally for the second definiton). Moreover, one can understand flatness as guaranteeing that closed subobjects for a Grothendieck topology generated by flat morphisms form a sheaf; this is part of why good quotient morphisms are flat in algebraic geometry, resp. open in topology. But I digress.
To show that projection morphisms are open, you have to use a different universal property of projection morphisms, which is where set theory comes into play. Namely, in the category of sets the projection morphism is also the folding map of an extensive copower, i.e. of an extensive coproduct of a set's worth copies of an object with itself.
What is happening is that in the category of sets there is an isomorphism $X\times Y\cong\bigsqcup_{x\in X}Y$. Explicitly, for each $x\in X$, we have an inclusion $j_x\colon Y\to X\times Y$ given by $y\mapsto(x,y)$. More categorically, if $\mathbf 1$ is the terminal object, the morphism $Y\to X\times Y$ are the pullbacks of $X\leftarrow X\times Y$ along each $\mathbf 1\to X$ corresponding to an element $x\in X$. These morphism $Y\to X\times Y$ form a colimiting cone.
Indeed, any family of set-functions $f_x\colon Y\to Z$ indexed by $x\in X$ corresponds, more or less by definition (depending on the set theory and how families are treated), to a function $f\colon X\times Y\to Z$ such that $f\circ j_x=f_x$. Again, this depends on how the set theory is set up, but you basically define $f\colon (x,y)\mapsto f_x(y)\in Z$.
In particular, the projection map $X\times Y\to Y$ now corresponds to the family of identity functions $Y\to Y$ indexed by $x\in X$, i.e. it is the folding map $\bigsqcup_{x\in X}Y\to Y$. Moreover, this copower is extensive: pullbacks of $\bigsqcup_{x\in X}Y\leftarrow Z$ along $j_x\colon Y\to\bigsqcup_{x\in X}Y$ determine the morphism $\bigsqcup_{x\in X}Y\leftarrow Y$.
With this out of the way, if we have a topological category over Set (topological spaces are an example, more generally any category of structured sets with coarsest and finiest structures is an example), then
we have that the folding map factors as $\bigsqcup_{x\in|X|}Y\cong|X|\times Y\to X\times Y\to Y$, where $|X|$ is the set $X$ with the finest topology, and $|X|\times Y\to X\times Y$ is induced from the unique $|X|\to X$.
There are now two steps. First, the folding map $p:\colon\bigsqcup_{x\in X}Y\cong |X|\times Y\to Y$ is open because $p^{-1}M\hookrightarrow \bar{p^{-1} M}\hookrightarrow p^{-1}\bar M\hookrightarrow|X|\times Y$ pulled back along each $j_x\colon Y\to |X|\times Y$ results in $M\cong id_Y^{-1}M\cong j_x^{-1}p^{-1}M\hookrightarrow j_x^{-1}\bar{p^{-1} M}\hookrightarrow j_x^{-1}p^{-1}\bar M\cong id_Y^{-1}\bar M\cong\bar M\hookrightarrow Y$, whence $j_x^{-1}\bar{p^{-1}M}\cong j_x^{-1}p^{-1}\bar M$ for each $x$, so the copower being extensive implies $\bar{p^{-1}M}\cong p^{-1}\bar M$.
Second, $|X|\times Y\to X\times Y$ is an instance of a morphism $A\to B$ for which we have an equality $|A|=|B|$ of the underlying sets, i.e. it is a morphism within a fiber of the topological functor to Set (and hence is a monomorphism). To complete the proof, it suffices for the closure operator to make these monomorphisms dense. Since such monomorphisms are stable under pullbacks, we see that they are stably dense. It is then a fact that canceling a stably dense morphism precedding an open morphism still yields an open morphism. Since this is long, you can find the short proof in the aforementioned Chapter III section 7 for the proof.

The property of openness of product projections is called semi-productivity by Dikranjan and Tholen. A slightly stronger notion called productivity is explored in Chapter 4 section 11 of their book. There they prove that if products are covered by sections, then every idempotent closure operator is productive, so in particular has open projections.
The notion of being covered by sections arises as follows. Pick two monomorphisms $m_1\colon M_1\hookrightarrow X_1$ and $m_2\colon M_2\hookrightarrow X_2$, and suppose a section $s\hookrightarrow X_1\to X_1\times X_2$ of $p_1\colon X_1\times X_2\to X_1$ (i.e. such that $p_1s=\mathrm{id}_{X_1}$) satisfies $s=(\mathrm{id}_{X_1}\times m_2)s'$ for some $s'\colon X_1\to X_1\times M_2$. Then $sm_1\colon M_1\hookrightarrow X_1\times X_2$ factors as $(m_1\times m_2)(\mathrm{id}_{X_1}\times m_2)s'$. Products in the category admit coverings by section if every $m_1\times m_2\colon M_1\times M_2\hookrightarrow X_1\times X_2$ is the union of the above-described sections.
Certainly the category of Sets has this property, which perhaps then implies that any category topological over Sets has it as well (this would be less weird than the argument above).
A: This is not a definite answer, but an attempt to explain why I do not expect that it is possible.

*

*Such a proof would require a characterization of open maps via some purely categorical description. I have never seen such a thing. This does not say too much, my knowledge is not encompassing.


*Assume that such a description exists. Then you could use this as a definition in any category. That is, there would be an abstract categorical concept of a certain type of morphisms which agrees with that of an open map in the category of topological spaces. I doubt that this is possible.


*If there exists a categorical concept of "open map", then I would expect that also a concept of "closed map" exists which should be similar. And if you can give a proof that the projections are categorically open, I would expect that they you can also prove that they are categorically closed. This is not true  in the category of topological spaces.
I know that this is a very weak argument, the categorical characterization of open  and closed maps may be so different that a proof is available only for the open case.
