$ \newcommand{\cat}[1]{\mathsf{#1}} $The title says it all. Let $ \cat J $ be a small category and let $ D $ be a $ \cat J $-indexed diagram of topological spaces. Suppose a limit $ (L,\tau\_L) $ of this diagram exists in the category of topological spaces (it exists, indeed, but that's not important here), and let $ {\left(\lambda\_j\colon (L,\tau\_L)\to Dj\right)}\_{j\in \cat J} $ be (the legs of) its corresponding limit cone.

Is it true that $ \tau\_L $ is the initial topology on the set $ L $ with respect to the continuous maps $ \lambda\_j\colon (L,\tau\_L)\to Dj $?

What I tried so far
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I tinkered with products just to get a feel for what I'm doing.

Let $ (A,\tau\_A) $ and $ (B,\tau\_B) $ be topological spaces. Let $ (P,\tau) $ be a categorical product of $ (A,\tau\_A) $ and $ (B,\tau\_B) $, and let $ \pi\_A\colon (P,\tau)\to (A,\tau\_A) $ and $ \pi\_B\colon (P,\tau)\to (B,\tau\_B) $ be (the legs of) its limit cone. Suppose $ \tau^\prime $ is a topology on the set $ P $ such that the set maps $ \pi\_A\colon P\to A $ and $ \pi\_B\colon P\to B $ upgrade to continuous functions $ \pi\_A\colon (P,\tau^\prime)\to (A,\tau\_A) $ and $ \pi\_B\colon (P,\tau^\prime)\to (B,\tau\_B) $ (I'm using the same names for the projections everywhere just to confuse you, or just because naming them differently is worse than the disease).

I need to prove that given $ U\in \tau $, then $ U\in \tau^\prime $. I thought that I could "pull back" $ U $ from $ (P,\tau) $ to $ (P,\tau^\prime) $ by the means of the continuous function $ \phi\colon (P,\tau^\prime)\to (P,\tau) $ whose existence is guaranteed by the universal property of products. If we are to follow this line of reasoning, we could well hope not only that $ \phi^{-1}(U) = U $, but that $ \phi $ has the identity function of $ P $ as the underlying set mapping.

From there I'm basically lost. Showing that the identity function $ 1\_P\colon P\to P $ give rise to a *continuous* function $ 1\_P\colon (P,\tau^\prime)\to (P,\tau) $ is the same as showing that $ \tau $ is coarser than $ \tau^\prime $, which is what I wanted to prove.