Let $\epsilon$ be a topos and let $f:Y\to X$ be a morphism. Let $M,N$ be two subobjects of $X$. Is it true that $f^*(M\cap N)=f^*(M)\cap f^*(N)$?

Question

Let $\epsilon$ be a topos and let $f:Y\longrightarrow X$ be a morphism in $\epsilon$. Now, let $M, N$ be two subobjects of $X$. Is it true that $f^*(M\cap N)=f^*(M)\cap f^*(N)$?

For $\epsilon=Set^{C^{{\rm op}}}$ where $C$ is a small category, I know it is true.

I tried to prove it for an arbitrary topos. But I couldn’t prove it and I am not sure that it is true.

Answer 1

This is true in any category whatsoever (as long as the category has binary pullbacks so the question even makes sense). This is a standard "pullback cube" argument, which is not to say that it's easy! At each of the following steps, you should verify that the diagram remains commutative.

First, form the pullback square defining $M \cap N$:

Next, paste on the pullback squares defining $f^*M$ and $f^* N$:

Next, paste on the pullback square defining $f^*(M \cap N)$:

Because right-hand face is a pullback, the maps $f^*(M \cap N) \to N$ and $f^*(M \cap N) \to Y$ induce a map $f^*(M \cap N) \to f^*N$:

Likewise, because the bottom face is a pullback, the maps $f^*(M \cap N) \to M$ and $f^*(M \cap N) \to Y$ induce a map $f^*(M \cap N) \to f^*(M)$:

Here is our completed "pullback cube":

Now, using the fact that the back, bottom, right, and "diagonal" (red) squares are pullbacks, prove that the front square is a pullback. This is a direct diagram chase, which @MaliceVidrine did for you :) This setup was important, though, because it tells us that the diagram actually commutes!

Answer 2

As pointed out in the comments, pullback has a left adjoint, so it preserves meets in the subobject lattice. But let's do the hairy diagram chasing proof just for fun.

Let $\alpha:f^*M\to M$, $\beta:f^*N\to N$, and $\gamma:f^*(M\cap N)\to M\cap N$ be the "top" edges of the pullbacks; let $m:M\to X$ and $n:N\to X$ be the respective inclusions; and let $i_M$ and $i_N$ be the inclusions of the intersection into $M,N$ respectively.

Now let's say that we have an object $D$ and $a:D\to f^*M$, $b:D\to f^*N$ with $f^*m\circ a=f^*n\circ b$. Then in particular we have $$m\circ\alpha\circ a=f\circ f^*m\circ a =f\circ f^*n\circ b=n\circ\beta\circ b$$ giving a unique morphism $j:D\to M\cap N$ with $i_M\circ j=\alpha\circ a$ and $i_N=\beta\circ b$.

But then $$m\circ i_M\circ j=m\circ\alpha\circ a= f\circ f^*m\circ a$$ giving a unique morphism $k:D\to f^*(M\cap N)$ such that $$f^*m\circ f^*i_M\circ k=f^*m\circ a$$ and $$\gamma\circ k=j.$$

Now notice that $f^*m\circ f^*i_M\circ k=f^*m\circ a$, since $f^*m$ is mono, means that $f^*i_M\circ k=a$. Notice also that our hypotheses on $a,b$ mean $$f^*n\circ f^*i_N\circ k=f^*m\circ f^*i_M\circ k=f^*m\circ a=f^*n\circ b$$ so that $f^*i_N\circ k=b$. Moreover, $k$ is unique with this property: if $f^*i_M\circ k=b=f^*i_M\circ k'$, then $k=k'$ since $f^*i_M$ is mono.

So from our hypotheses on $D,a,b$ there is a unique $k:D\to f^*(M\cap N)$ with $f^*i_M\circ k=a$ and $f^*i_N\circ k=b$; but this says precisely that $f^*(M\cap N)$ is a pullback of $f^*m$ and $f^*n$, making it $f^*M\cap f^*N$.

Answer 3

The standard proof in $Set$ that $f^*(M \cap N) = f^*(M) \cap f^*(N)$ will translate into the internal language of a topos just fine.

The interpretation of this standard proof would go like:

Let $U$ be any test object of the topos and $x : U \to X$ any morphism. Then \begin{align*} x\mathrm{~factors~through~}f^*(M\cap N) & \iff f\circ x\mathrm{~factors~through~}M\cap N \\ & \iff f\circ x\mathrm{~factors~through~}M \wedge f\circ x \mathrm{~factors~through~}N \\ & \iff x\mathrm{~factors~through~}f^*(M) \wedge x\mathrm{~factors~through~}f^*(N) \\ & \iff x\mathrm{~factors~through~}f^*(M) \cap f^*(N). \end{align*} Now, applying this with $U := f^*(M \cap N)$ and $x$ the inclusion, we see $f^*(M\cap N)$ factors through $f^*(M) \cap f^*(N)$; and similarly, in the other direction, $f^*(M) \cap f^*(N)$ factors through $f^*(M\cap N)$. It follows that $f^*(M\cap N) = f^*(M) \cap f^*(N)$ as subobjects of $X$.

(Note that the interpretation of the standard proof that $f^*(M \cup N) = f^*(M) \cup f^*(N)$ would be a little more involved. Here, we would need to use that $x : U \to X$ factors through $S \cup T$ if and only if there exist $\pi_1 : V_1 \to U$ and $\pi_2 : V_2 \to U$ such that $\pi_1 \sqcup \pi_2 : V_1 \sqcup V_2 \to U$ is an epimorphism; $\pi_1^*(x) = x \circ \pi_1$ factors through $S$; and $\pi_2^*(x) = x \circ \pi_2$ factors through $T$.)