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Let $(X_j,\tau _j), j\in I$ be topological spaces, where $I$ is a set. Let $$(Y,\tau ^\times) := \left (\prod_{j\in I}X_j,\tau ^\times\right ) $$ be the product topology, where $\tau ^\times$ is the weakest topology s.t the projection mapping $\pi _j : (Y,\tau ^\times)\to (X_j,\tau _j)$ is continuous.

Let $(Z,\tau _Z)$ be a topological space and $f:Z\to Y$

Show that $f$ is continuous if and only if for every $j\in I$, $\pi _jf$ is continous.

The forward implication holds because composition of continous functions is continous.

Having trouble with reverse implication. Here in the accepted answer they say the sets of the form $$\bigcap_{k=1}^n \pi _j^{-1}(V_{j_k}), V_{j_k}\in\tau _{j_k}, j_1,\ldots ,j_k\in I, n\in\mathbb{N} $$ are a basis on the product topology which is indeed, easy to see. What isn't so clear is where exactly they use the continuity of $\pi _jf$.

To show $f$ is continous, let $V\in\tau ^\times$. Need to show that $f^{-1}(V)\in\tau _Z$. Consider the collection $$\tau = \lbrace U\subset Y :f^{-1}(U)\in\tau _Z \rbrace$$ It is very straightforward to verify $(Y,\tau)$ is a topological space. IF $\pi _j : (Y,\tau)\to (X_j,\tau _j)$ is continous, then by definition, we would have $\tau ^\times\subset\tau$, implying $f^{-1}(V)\in\tau _Z$.

Drawing fairly blank on continuity of $\pi _j$ w.r.t to this new topology $\tau$ and still perplexed about how to utilize the premise of continuity of $\pi _jf$. It only provides that the preimage of $V\in\tau _j$ is open in $Z$ and it's not true, in general, that if the composition is continous, its components are continous.

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  • $\begingroup$ Did you know that inverses preserve intersection? $f^{-1}(\cap_i S_i)=\cap_i f^{-1}(S_i)$. You've described all the other components you need in the question. $\endgroup$ Nov 22 '16 at 22:05
  • $\begingroup$ @enthdegree What are you getting at? We pick $V\in\tau ^\times$, where $V\overset{?}=\prod_{j\in I} U_j$. Then by assumption for every $j\in I$ $(\pi _jf)^{-1}(U_j)\in\tau _Z$. Now $f^{-1}\left (\prod U_j\right )\overset{?}=(\pi _jf)^{-1}(\bigcap_{j\in I}U_j)$. If $I$ was finite, I would agree, but it need not be. I might be completely off the mark about what you meant, too. $\endgroup$
    – AlvinL
    Nov 22 '16 at 22:28
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You have seen that if $S\in \tau^\times$ then $S$ can be expanded $$S=\cup_{\lambda\in \Lambda}\left(\cap_{j\in I_\lambda} \pi_j^{-1}(V^\lambda_{j})\right), \quad I_\lambda \text{ finite},\quad V^{\lambda}_{j}\in \tau_j$$

We know that in general, (disregarding topology) for any collection $\{U_\gamma\}_{\gamma\in \Gamma}$ then $f^{-1}(\cap_\gamma U_\gamma)=\cap_\gamma f^{-1}(U_\gamma)$ and $f^{-1}(\cup_\gamma U_\gamma)=\cup_\gamma f^{-1}(U_\gamma).$ So certainly we can write:

$$f^{-1}(S)=\cup_{\lambda\in \Lambda}\left[\cap_{j\in I_\lambda} f^{-1}\left(\pi_j^{-1}(V^\lambda_{j})\right)\right].$$

By assumption each $f^{-1}\left(\pi_j^{-1}(V_j^\lambda)\right)\in \tau_Z$ so we conclude $f^{-1}(S)$ is also.

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  • $\begingroup$ Oh right, I had my doubts about the preimage property you were claiming, but I verified that $$x\in (fg)^{-1}(A)\iff x\in g^{-1}(f^{-1}(A)) $$ $\endgroup$
    – AlvinL
    Nov 23 '16 at 8:28

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